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bg-background","children":["$","$L39",null,{"backHref":"/a-level/maths/wjec/questions","basePath":"/a-level/maths/wjec/sequences-and-series/questions","formulaUrl":"https://www.wjec.co.uk/media/4jfdytcz/ss-w27-18-e.pdf","initialIndex":80,"initialQuestionWindow":[{"id":"99447d6d-aed8-474f-b6d1-01a7abc0d1b0","revisionId":"de88fd87-e160-49d3-baa3-5c9aa4e709cc","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"2373e137-f4c5-45c1-a32a-569fa5852934","label":"a","content":"Work out the first three terms in the binomial expansion of $\\frac{1}{(2-5x)^2}$","markscheme":"- Writes as $(2-5x)^{-2} = 2^{-2}(1 - \\frac{5}{2}x)^{-2} = \\frac{1}{4}(1 - \\frac{5}{2}x)^{-2}$ M1\n- Expands: $\\frac{1}{4} [ 1 + (-2)(-\\frac{5}{2}x) + \\frac{(-2)(-3)}{2}(-\\frac{5}{2}x)^2 ]$ M1\n- Simplifies terms inside the bracket: $1 + 5x + \\frac{75}{4}x^2$ A1\n- Final answer: $\\frac{1}{4} + \\frac{5}{4}x + \\frac{75}{16}x^2$ A1","marks":4,"order":0,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Correctly identifies factor $\\frac{1}{4}$ and writes expression in form $(1+u)^{-2}$.","code":"M"},{"id":"p2","marks":1,"description":"Attempts binomial expansion with $n=-2$ and $u = -\\frac{5}{2}x$.","code":"M"},{"id":"p3","marks":1,"description":"Obtains correct terms inside the bracket $1 + 5x + \\frac{75}{4}x^2$.","code":"A"},{"id":"p4","marks":1,"description":"Correct final answer $\\frac{1}{4} + \\frac{5}{4}x + \\frac{75}{16}x^2$.","code":"A"}],"finalAnswers":["\\frac{1}{4} + \\frac{5}{4}x + \\frac{75}{16}x^2","0.25 + 1.25x + 4.6875x^2"],"generatedAt":"2026-05-06T14:04:06.821Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"factoring_and_expansion","label":"Factoring and Standard Binomial Formula","steps":[{"type":"text","order":0,"title":"Rewrite using negative indices","content":"First, we rewrite the fraction $\\frac{1}{(2-5x)^2}$ using a negative index:\n\n$$\\frac{1}{(2-5x)^2} = (2-5x)^{-2}$$\n\nThis puts our expression in the form $(a + bx)^n$, where $n = -2$.","highlights":[{"text":"Work out the first three terms in the binomial expansion of $\\frac{1}{(2-5x)^2}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\left(\\htmlData{anchor=const}{\\color{@sky}{2}} - 5x\\right)^{-2} = \\htmlData{anchor=const-pow}{\\color{@sky}{2^{-2}}}\\left(1 - \\htmlData{anchor=u-term}{\\color{@amber}{\\frac{5}{2}x}}\\right)^{-2}$"},{"latex":"$$= \\htmlData{anchor=coeff}{\\color{@sky}{\\frac{1}{4}}}\\left(1 - \\htmlData{anchor=u-term-simp}{\\color{@amber}{2.5x}}\\right)^{-2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"const"},{"color":"sky","style":"text-color","anchor":"const-pow"},{"color":"sky","style":"text-color","anchor":"coeff"}],"connections":[{"to":"const-pow","from":"const","color":"sky","label":"factor out base","style":"solid"},{"to":"coeff","from":"const-pow","color":"sky","label":"evaluate","style":"solid"},{"to":"u-term-simp","from":"u-term","color":"amber","label":"simplify fraction","style":"solid"}]},"order":1,"title":"Factor out the constant","description":"To use the standard binomial expansion formula, we must factor out the leading constant so that the bracket starts with $1$:\n\n$$(2-5x)^{-2} = \\left[2\\left(1 - \\frac{5}{2}x\\right)\\right]^{-2} = 2^{-2}\\left(1 - \\frac{5}{2}x\\right)^{-2} = \\frac{1}{4}\\left(1 - \\frac{5}{2}x\\right)^{-2}$$\n\nThis earns the first method mark **(M1)**."},{"type":"text","order":2,"title":"Apply binomial formula","content":"Next, we apply the standard binomial expansion formula:\n\n$$(1 + u)^n = 1 + nu + \\frac{n(n-1)}{2!}u^2 + \\dots$$\n\nIn our case, we have $n = -2$ and $u = -\\frac{5}{2}x$ (or $-2.5x$). Substituting these values into the bracket, we get:\n\n$$\\left(1 - \\frac{5}{2}x\\right)^{-2} = 1 + (-2)\\left(-\\frac{5}{2}x\\right) + \\frac{(-2)(-3)}{2!}\\left(-\\frac{5}{2}x\\right)^2 + \\dots$$\n\nThis step earns the second method mark **(M1)**."},{"type":"text","order":3,"title":"Simplify terms in bracket","content":"Let's simplify each term inside the square brackets:\n\n* **First term:** $1$\n* **Second term:** $(-2)\\left(-\\frac{5}{2}x\\right) = 5x$\n* **Third term:** $\\frac{6}{2}\\left(\\frac{25}{4}x^2\\right) = 3\\left(\\frac{25}{4}x^2\\right) = \\frac{75}{4}x^2$\n\nPutting these together gives:\n\n$$1 + 5x + \\frac{75}{4}x^2$$\n\nThis earns an accuracy mark **(A1)**."},{"type":"text","order":4,"title":"Multiply by constant factor","content":"Finally, we multiply the simplified series inside the brackets by the constant factor $\\frac{1}{4}$:\n\n$$\\frac{1}{4}\\left(1 + 5x + \\frac{75}{4}x^2\\right) = \\frac{1}{4} + \\frac{5}{4}x + \\frac{75}{16}x^2$$\n\nThis gives our final expanded expression, earning the final accuracy mark **(A1)**."}]},{"id":"direct_general_binomial","label":"Direct General Binomial Formula","steps":[{"type":"text","order":0,"title":"Rewrite the expression","content":"First, let's write the given algebraic fraction as a single expression with a negative power:\n\n$$\\frac{1}{(2-5x)^2} = (2-5x)^{-2}$$\n\nThis is now in the form $(a+b)^n$, where we can directly apply the general binomial expansion formula."},{"type":"text","order":1,"title":"State the formula","content":"The general binomial expansion formula for any real power $n$ of the binom $(a + b)^n$ is:\n\n$$(a + b)^n = a^n + n a^{n-1}b + \\frac{n(n-1)}{2!} a^{n-2}b^2 + \\dots$$\n\nBy matching terms in our expression $(2-5x)^{-2}$ to the general formula, we identify:\n* $a = 2$\n* $b = -5x$\n* $n = -2$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$a = \\htmlData{anchor=a-val}{\\color{@sky}{2}}, \\; b = \\htmlData{anchor=b-val}{\\color{@amber}{-5x}}, \\; n = \\htmlData{anchor=n-val}{\\color{@purple}{-2}}$"},{"latex":"$$(\\color{@sky}{a} + \\color{@amber}{b})^{\\color{@purple}{n}} \\approx \\htmlData{anchor=term1}{\\color{@sky}{a}^{\\color{@purple}{n}}} + \\htmlData{anchor=term2}{\\color{@purple}{n}\\color{@sky}{a}^{\\color{@purple}{n}-1}\\color{@amber}{b}} + \\htmlData{anchor=term3}{\\frac{\\color{@purple}{n}(\\color{@purple}{n}-1)}{2} \\color{@sky}{a}^{\\color{@purple}{n}-2}\\color{@amber}{b}^2}$"},{"latex":"$$(\\color{@sky}{2} \\color{@amber}{- 5x})^{\\color{@purple}{-2}} \\approx \\color{@sky}{2}^{\\color{@purple}{-2}} + (\\color{@purple}{-2})\\color{@sky}{2}^{\\color{@purple}{-3}}(\\color{@amber}{-5x}) + \\frac{(\\color{@purple}{-2})(\\color{@purple}{-3})}{2}\\color{@sky}{2}^{\\color{@purple}{-4}}(\\color{@amber}{-5x})^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"amber","style":"text-color","anchor":"b-val"},{"color":"purple","style":"text-color","anchor":"n-val"}],"connections":[{"to":"term1","from":"a-val","color":"sky","label":"a","style":"solid"},{"to":"term2","from":"b-val","color":"amber","label":"b","style":"solid"},{"to":"term3","from":"n-val","color":"purple","label":"n","style":"solid"}]},"order":2,"title":"Substitute into the formula","description":"We substitute our values $a=2$, $b=-5x$, and $n=-2$ directly into the general binomial series expansion."},{"type":"text","order":3,"title":"Evaluate the terms","content":"Now, let's calculate each of the three terms individually:\n\n* **First term ($a^n$):**\n $$2^{-2} = \\frac{1}{2^2} = \\frac{1}{4}$$\n\n* **Second term ($n a^{n-1}b$):**\n $$(-2) \\cdot 2^{-3} \\cdot (-5x) = (-2) \\cdot \\frac{1}{8} \\cdot (-5x) = \\frac{10}{8}x = \\frac{5}{4}x$$\n\n* **Third term ($\\frac{n(n-1)}{2!} a^{n-2}b^2$):**\n $$\\frac{(-2)(-3)}{2} \\cdot 2^{-4} \\cdot (-5x)^2 = 3 \\cdot \\frac{1}{16} \\cdot (25x^2) = \\frac{75}{16}x^2$$"},{"type":"text","order":4,"title":"Write final answer","content":"Combining the evaluated terms gives the final three terms of our binomial expansion:\n\n$$\\frac{1}{(2-5x)^2} \\approx \\frac{1}{4} + \\frac{5}{4}x + \\frac{75}{16}x^2$$"}]},{"id":"maclaurin_series","label":"Maclaurin Series Expansion","steps":[{"type":"text","order":0,"title":"Identify the Maclaurin formula","content":"The Maclaurin series expansion of a function $f(x)$ up to the quadratic term is given by:\n\n$$f(x) ≈ f(0) + f'(0)x + \\frac{f''(0)}{2!}x^2$$\n\nTo find the first three terms of the expansion for $f(x) = (2-5x)^{-2}$, we need to calculate $f(0)$, the first derivative $f'(0)$, and the second derivative $f''(0)$."},{"type":"text","order":1,"title":"Evaluate at $x = 0$","content":"Let's first find the value of the function when $x = 0$:\n\n$$f(x) = (2 - 5x)^{-2}$$\n\nSubstituting $x = 0$:\n\n$$\\begin{aligned} f(0) &= (2 - 5(0))^{-2} \\\\ &= 2^{-2} \\\\ &= \\frac{1}{4} \\end{aligned}$$"},{"type":"text","order":2,"title":"Find the first derivative","content":"Using the chain rule, we find the first derivative $f'(x)$:\n\n$$f'(x) = -2(2 - 5x)^{-3} \\cdot (-5)$$\n\n$$\\begin{aligned} f'(x) &= 10(2 - 5x)^{-3} \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Substitute } x = 0 \\text{ into the first derivative:}$"},{"latex":"$$f'(0) = 10(2 - 5(\\htmlData{anchor=x-val}{0}))^{-3}$"},{"latex":"$$f'(0) = 10(2)^{-3}$"},{"latex":"$$f'(0) = \\htmlData{anchor=num-part}{10} \\cdot \\htmlData{anchor=denom-part}{\\frac{1}{8}}$"},{"latex":"$$f'(0) = \\frac{10}{8} = \\frac{5}{4}$"}],"highlights":[{"color":"amber","style":"box","anchor":"x-val"}],"connections":[{"to":"denom-part","from":"num-part","color":"sky","label":"multiply","style":"solid"}]},"order":3,"title":"Evaluate $f'(0)$","description":"Substituting $x = 0$ allows us to find the first derivative's value, which simplifies to $\\frac{5}{4}$."},{"type":"text","order":4,"title":"Find the second derivative","content":"Now we differentiate $f'(x) = 10(2 - 5x)^{-3}$ again using the chain rule to find $f''(x)$:\n\n$$f''(x) = 10 \\cdot (-3)(2 - 5x)^{-4} \\cdot (-5)$$\n\n$$\\begin{aligned} f''(x) &= -30(2 - 5x)^{-4} \\cdot (-5) \\\\ f''(x) &= 150(2 - 5x)^{-4} \\end{aligned}$$"},{"type":"text","order":5,"title":"Evaluate $f''(0)$","content":"Substitute $x = 0$ into the second derivative:\n\n$$\\begin{aligned} f''(0) &= 150(2 - 5(0))^{-4} \\\\ &= 150(2)^{-4} \\\\ &= 150 \\cdot \\frac{1}{16} \\\\ &= \\frac{150}{16} \\\\ &= \\frac{75}{8} \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Substitute } \\htmlData{anchor=f0}{\\frac{1}{4}}, \\; \\htmlData{anchor=f1}{\\frac{5}{4}}, \\text{ and } \\htmlData{anchor=f2}{\\frac{75}{8}} \\text{ into the Maclaurin series:}$"},{"latex":"$$f(x) \\approx \\htmlData{anchor=term0}{\\frac{1}{4}} + \\htmlData{anchor=term1}{\\left(\\frac{5}{4}\\right)}x + \\frac{\\htmlData{anchor=term2}{\\frac{75}{8}}}{2!}x^2$"},{"latex":"$$f(x) \\approx \\frac{1}{4} + \\frac{5}{4}x + \\frac{75}{16}x^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"f0"},{"color":"amber","style":"text-color","anchor":"f1"},{"color":"purple","style":"text-color","anchor":"f2"}],"connections":[{"to":"term0","from":"f0","color":"sky","label":null,"style":"solid"},{"to":"term1","from":"f1","color":"amber","label":null,"style":"solid"},{"to":"term2","from":"f2","color":"purple","label":null,"style":"solid"}]},"order":6,"title":"Substitute into the formula","description":"Applying the values to the Maclaurin formula $f(0) + f'(0)x + \\frac{f''(0)}{2}x^2$ gives the first three terms of our series expansion."}]}],"generatedAt":"2026-06-09T02:26:17.461Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"bc86cc9d-c267-4347-b8eb-07f3053bd96b","label":"b","content":"Write down the range of values of $x$ for which the expansion is valid.","markscheme":"- $|-\\frac{5}{2}x| < 1 \\Rightarrow |x| < \\frac{2}{5}$ B1","marks":1,"order":1,"markingRubric":{"totalMarks":1,"points":[{"id":"p1","marks":1,"description":"Identifies $|x| < \\frac{2}{5}$ or $0.4$.","code":"B"}],"finalAnswers":["|x| < \\frac{2}{5}","|x| < 0.4","-0.4 < x < 0.4"],"generatedAt":"2026-05-06T14:04:06.821Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"binomial_convergence_condition","label":"Binomial Convergence Condition","steps":[{"type":"text","order":0,"title":"Identify the convergence condition","content":"For any binomial expansion of the form $(1 + kx)^n$ where $n$ is not a positive integer, the infinite series converges (is valid) if and only if the absolute value of the variable term is less than $1$:\n\n$$|kx| < 1$$"},{"type":"text","order":1,"title":"Substitute the variable term","content":"In our expansion, the term in the position of $kx$ is $-\\frac{5}{2}x$.\n\nSubstituting this into the convergence condition gives:\n\n$$\\left|-\\frac{5}{2}x\\right| < 1$$"},{"type":"text","order":2,"title":"Solve the inequality","content":"Since the absolute value of a negative coefficient is positive, we can simplify this expression:\n\n$$\\frac{5}{2}|x| < 1$$\n\nMultiplying both sides by $\\frac{2}{5}$ gives:\n\n$$|x| < \\frac{2}{5}$$"},{"type":"text","order":3,"title":"State the final range","content":"Thus, the expansion is valid for $|x| < \\frac{2}{5}$ (or $-0.4 < x < 0.4$)."}]}],"generatedAt":"2026-06-09T02:26:52.768Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"2373e137-f4c5-45c1-a32a-569fa5852934:factoring_and_expansion","label":"(a) Factoring and Standard Binomial Formula","steps":[{"type":"text","order":0,"title":"Rewrite using negative indices","content":"First, we rewrite the fraction $\\frac{1}{(2-5x)^2}$ using a negative index:\n\n$$\\frac{1}{(2-5x)^2} = (2-5x)^{-2}$$\n\nThis puts our expression in the form $(a + bx)^n$, where $n = -2$.","highlights":[{"text":"Work out the first three terms in the binomial expansion of $\\frac{1}{(2-5x)^2}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\left(\\htmlData{anchor=const}{\\color{@sky}{2}} - 5x\\right)^{-2} = \\htmlData{anchor=const-pow}{\\color{@sky}{2^{-2}}}\\left(1 - \\htmlData{anchor=u-term}{\\color{@amber}{\\frac{5}{2}x}}\\right)^{-2}$"},{"latex":"$$= \\htmlData{anchor=coeff}{\\color{@sky}{\\frac{1}{4}}}\\left(1 - \\htmlData{anchor=u-term-simp}{\\color{@amber}{2.5x}}\\right)^{-2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"const"},{"color":"sky","style":"text-color","anchor":"const-pow"},{"color":"sky","style":"text-color","anchor":"coeff"}],"connections":[{"to":"const-pow","from":"const","color":"sky","label":"factor out base","style":"solid"},{"to":"coeff","from":"const-pow","color":"sky","label":"evaluate","style":"solid"},{"to":"u-term-simp","from":"u-term","color":"amber","label":"simplify fraction","style":"solid"}]},"order":1,"title":"Factor out the constant","description":"To use the standard binomial expansion formula, we must factor out the leading constant so that the bracket starts with $1$:\n\n$$(2-5x)^{-2} = \\left[2\\left(1 - \\frac{5}{2}x\\right)\\right]^{-2} = 2^{-2}\\left(1 - \\frac{5}{2}x\\right)^{-2} = \\frac{1}{4}\\left(1 - \\frac{5}{2}x\\right)^{-2}$$\n\nThis earns the first method mark **(M1)**."},{"type":"text","order":2,"title":"Apply binomial formula","content":"Next, we apply the standard binomial expansion formula:\n\n$$(1 + u)^n = 1 + nu + \\frac{n(n-1)}{2!}u^2 + \\dots$$\n\nIn our case, we have $n = -2$ and $u = -\\frac{5}{2}x$ (or $-2.5x$). Substituting these values into the bracket, we get:\n\n$$\\left(1 - \\frac{5}{2}x\\right)^{-2} = 1 + (-2)\\left(-\\frac{5}{2}x\\right) + \\frac{(-2)(-3)}{2!}\\left(-\\frac{5}{2}x\\right)^2 + \\dots$$\n\nThis step earns the second method mark **(M1)**."},{"type":"text","order":3,"title":"Simplify terms in bracket","content":"Let's simplify each term inside the square brackets:\n\n* **First term:** $1$\n* **Second term:** $(-2)\\left(-\\frac{5}{2}x\\right) = 5x$\n* **Third term:** $\\frac{6}{2}\\left(\\frac{25}{4}x^2\\right) = 3\\left(\\frac{25}{4}x^2\\right) = \\frac{75}{4}x^2$\n\nPutting these together gives:\n\n$$1 + 5x + \\frac{75}{4}x^2$$\n\nThis earns an accuracy mark **(A1)**."},{"type":"text","order":4,"title":"Multiply by constant factor","content":"Finally, we multiply the simplified series inside the brackets by the constant factor $\\frac{1}{4}$:\n\n$$\\frac{1}{4}\\left(1 + 5x + \\frac{75}{4}x^2\\right) = \\frac{1}{4} + \\frac{5}{4}x + \\frac{75}{16}x^2$$\n\nThis gives our final expanded expression, earning the final accuracy mark **(A1)**."}]},{"id":"2373e137-f4c5-45c1-a32a-569fa5852934:direct_general_binomial","label":"(a) Direct General Binomial Formula","steps":[{"type":"text","order":0,"title":"Rewrite the expression","content":"First, let's write the given algebraic fraction as a single expression with a negative power:\n\n$$\\frac{1}{(2-5x)^2} = (2-5x)^{-2}$$\n\nThis is now in the form $(a+b)^n$, where we can directly apply the general binomial expansion formula."},{"type":"text","order":1,"title":"State the formula","content":"The general binomial expansion formula for any real power $n$ of the binom $(a + b)^n$ is:\n\n$$(a + b)^n = a^n + n a^{n-1}b + \\frac{n(n-1)}{2!} a^{n-2}b^2 + \\dots$$\n\nBy matching terms in our expression $(2-5x)^{-2}$ to the general formula, we identify:\n* $a = 2$\n* $b = -5x$\n* $n = -2$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$a = \\htmlData{anchor=a-val}{\\color{@sky}{2}}, \\; b = \\htmlData{anchor=b-val}{\\color{@amber}{-5x}}, \\; n = \\htmlData{anchor=n-val}{\\color{@purple}{-2}}$"},{"latex":"$$(\\color{@sky}{a} + \\color{@amber}{b})^{\\color{@purple}{n}} \\approx \\htmlData{anchor=term1}{\\color{@sky}{a}^{\\color{@purple}{n}}} + \\htmlData{anchor=term2}{\\color{@purple}{n}\\color{@sky}{a}^{\\color{@purple}{n}-1}\\color{@amber}{b}} + \\htmlData{anchor=term3}{\\frac{\\color{@purple}{n}(\\color{@purple}{n}-1)}{2} \\color{@sky}{a}^{\\color{@purple}{n}-2}\\color{@amber}{b}^2}$"},{"latex":"$$(\\color{@sky}{2} \\color{@amber}{- 5x})^{\\color{@purple}{-2}} \\approx \\color{@sky}{2}^{\\color{@purple}{-2}} + (\\color{@purple}{-2})\\color{@sky}{2}^{\\color{@purple}{-3}}(\\color{@amber}{-5x}) + \\frac{(\\color{@purple}{-2})(\\color{@purple}{-3})}{2}\\color{@sky}{2}^{\\color{@purple}{-4}}(\\color{@amber}{-5x})^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"amber","style":"text-color","anchor":"b-val"},{"color":"purple","style":"text-color","anchor":"n-val"}],"connections":[{"to":"term1","from":"a-val","color":"sky","label":"a","style":"solid"},{"to":"term2","from":"b-val","color":"amber","label":"b","style":"solid"},{"to":"term3","from":"n-val","color":"purple","label":"n","style":"solid"}]},"order":2,"title":"Substitute into the formula","description":"We substitute our values $a=2$, $b=-5x$, and $n=-2$ directly into the general binomial series expansion."},{"type":"text","order":3,"title":"Evaluate the terms","content":"Now, let's calculate each of the three terms individually:\n\n* **First term ($a^n$):**\n $$2^{-2} = \\frac{1}{2^2} = \\frac{1}{4}$$\n\n* **Second term ($n a^{n-1}b$):**\n $$(-2) \\cdot 2^{-3} \\cdot (-5x) = (-2) \\cdot \\frac{1}{8} \\cdot (-5x) = \\frac{10}{8}x = \\frac{5}{4}x$$\n\n* **Third term ($\\frac{n(n-1)}{2!} a^{n-2}b^2$):**\n $$\\frac{(-2)(-3)}{2} \\cdot 2^{-4} \\cdot (-5x)^2 = 3 \\cdot \\frac{1}{16} \\cdot (25x^2) = \\frac{75}{16}x^2$$"},{"type":"text","order":4,"title":"Write final answer","content":"Combining the evaluated terms gives the final three terms of our binomial expansion:\n\n$$\\frac{1}{(2-5x)^2} \\approx \\frac{1}{4} + \\frac{5}{4}x + \\frac{75}{16}x^2$$"}]},{"id":"2373e137-f4c5-45c1-a32a-569fa5852934:maclaurin_series","label":"(a) Maclaurin Series Expansion","steps":[{"type":"text","order":0,"title":"Identify the Maclaurin formula","content":"The Maclaurin series expansion of a function $f(x)$ up to the quadratic term is given by:\n\n$$f(x) ≈ f(0) + f'(0)x + \\frac{f''(0)}{2!}x^2$$\n\nTo find the first three terms of the expansion for $f(x) = (2-5x)^{-2}$, we need to calculate $f(0)$, the first derivative $f'(0)$, and the second derivative $f''(0)$."},{"type":"text","order":1,"title":"Evaluate at $x = 0$","content":"Let's first find the value of the function when $x = 0$:\n\n$$f(x) = (2 - 5x)^{-2}$$\n\nSubstituting $x = 0$:\n\n$$\\begin{aligned} f(0) &= (2 - 5(0))^{-2} \\\\ &= 2^{-2} \\\\ &= \\frac{1}{4} \\end{aligned}$$"},{"type":"text","order":2,"title":"Find the first derivative","content":"Using the chain rule, we find the first derivative $f'(x)$:\n\n$$f'(x) = -2(2 - 5x)^{-3} \\cdot (-5)$$\n\n$$\\begin{aligned} f'(x) &= 10(2 - 5x)^{-3} \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Substitute } x = 0 \\text{ into the first derivative:}$"},{"latex":"$$f'(0) = 10(2 - 5(\\htmlData{anchor=x-val}{0}))^{-3}$"},{"latex":"$$f'(0) = 10(2)^{-3}$"},{"latex":"$$f'(0) = \\htmlData{anchor=num-part}{10} \\cdot \\htmlData{anchor=denom-part}{\\frac{1}{8}}$"},{"latex":"$$f'(0) = \\frac{10}{8} = \\frac{5}{4}$"}],"highlights":[{"color":"amber","style":"box","anchor":"x-val"}],"connections":[{"to":"denom-part","from":"num-part","color":"sky","label":"multiply","style":"solid"}]},"order":3,"title":"Evaluate $f'(0)$","description":"Substituting $x = 0$ allows us to find the first derivative's value, which simplifies to $\\frac{5}{4}$."},{"type":"text","order":4,"title":"Find the second derivative","content":"Now we differentiate $f'(x) = 10(2 - 5x)^{-3}$ again using the chain rule to find $f''(x)$:\n\n$$f''(x) = 10 \\cdot (-3)(2 - 5x)^{-4} \\cdot (-5)$$\n\n$$\\begin{aligned} f''(x) &= -30(2 - 5x)^{-4} \\cdot (-5) \\\\ f''(x) &= 150(2 - 5x)^{-4} \\end{aligned}$$"},{"type":"text","order":5,"title":"Evaluate $f''(0)$","content":"Substitute $x = 0$ into the second derivative:\n\n$$\\begin{aligned} f''(0) &= 150(2 - 5(0))^{-4} \\\\ &= 150(2)^{-4} \\\\ &= 150 \\cdot \\frac{1}{16} \\\\ &= \\frac{150}{16} \\\\ &= \\frac{75}{8} \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Substitute } \\htmlData{anchor=f0}{\\frac{1}{4}}, \\; \\htmlData{anchor=f1}{\\frac{5}{4}}, \\text{ and } \\htmlData{anchor=f2}{\\frac{75}{8}} \\text{ into the Maclaurin series:}$"},{"latex":"$$f(x) \\approx \\htmlData{anchor=term0}{\\frac{1}{4}} + \\htmlData{anchor=term1}{\\left(\\frac{5}{4}\\right)}x + \\frac{\\htmlData{anchor=term2}{\\frac{75}{8}}}{2!}x^2$"},{"latex":"$$f(x) \\approx \\frac{1}{4} + \\frac{5}{4}x + \\frac{75}{16}x^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"f0"},{"color":"amber","style":"text-color","anchor":"f1"},{"color":"purple","style":"text-color","anchor":"f2"}],"connections":[{"to":"term0","from":"f0","color":"sky","label":null,"style":"solid"},{"to":"term1","from":"f1","color":"amber","label":null,"style":"solid"},{"to":"term2","from":"f2","color":"purple","label":null,"style":"solid"}]},"order":6,"title":"Substitute into the formula","description":"Applying the values to the Maclaurin formula $f(0) + f'(0)x + \\frac{f''(0)}{2}x^2$ gives the first three terms of our series expansion."}]},{"id":"bc86cc9d-c267-4347-b8eb-07f3053bd96b:binomial_convergence_condition","label":"(b) Binomial Convergence Condition","steps":[{"type":"text","order":0,"title":"Identify the convergence condition","content":"For any binomial expansion of the form $(1 + kx)^n$ where $n$ is not a positive integer, the infinite series converges (is valid) if and only if the absolute value of the variable term is less than $1$:\n\n$$|kx| < 1$$"},{"type":"text","order":1,"title":"Substitute the variable term","content":"In our expansion, the term in the position of $kx$ is $-\\frac{5}{2}x$.\n\nSubstituting this into the convergence condition gives:\n\n$$\\left|-\\frac{5}{2}x\\right| < 1$$"},{"type":"text","order":2,"title":"Solve the inequality","content":"Since the absolute value of a negative coefficient is positive, we can simplify this expression:\n\n$$\\frac{5}{2}|x| < 1$$\n\nMultiplying both sides by $\\frac{2}{5}$ gives:\n\n$$|x| < \\frac{2}{5}$$"},{"type":"text","order":3,"title":"State the final range","content":"Thus, the expansion is valid for $|x| < \\frac{2}{5}$ (or $-0.4 < x < 0.4$)."}]}],"generatedAt":"2026-06-09T02:27:03.441Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"aff5e867-ea3b-4a26-9a7c-0af68ae1a615","revisionId":"2f05522f-b3fc-475a-aaa7-7957d69a51fd","questionNumber":null,"questionType":"LA","specification":"The first three terms of an arithmetic series are $(p+5)$, $(3p-1)$ and $(4p+3)$ respectively.","difficulty":1,"difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"739eff19-09b7-4564-bf50-3192c3eff459","label":"a","content":"Find the value of the constant $p$.","markscheme":"- Use common difference property: $(3p-1) - (p+5) = (4p+3) - (3p-1)$ M1\n- $2p - 6 = p + 4 \\implies p = 10$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Sets up an equation using the common difference property $(u_2 - u_1 = u_3 - u_2)$.","code":"M"},{"id":"p2","marks":1,"description":"Correctly solves the linear equation to find $p = 10$.","code":"A"}],"finalAnswers":["10"],"generatedAt":"2026-05-06T14:05:22.840Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"common_difference","label":"Equating Common Differences","steps":[{"type":"text","order":0,"title":"Identify the terms","content":"The question gives us the first three terms of an arithmetic series:\n\n* First term: $u_1 = p + 5$\n* Second term: $u_2 = 3p - 1$\n* Third term: $u_3 = 4p + 3$","highlights":[{"text":"(p+5)","location":"specification"},{"text":"(3p-1)","location":"specification"},{"text":"(4p+3)","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$d = \\text{Term } 2 - \\text{Term } 1 = \\text{Term } 3 - \\text{Term } 2$$"},{"latex":"$$$\\htmlData{anchor=diff-1-lhs}{\\color{@sky}{(3p - 1) - (p + 5)}} = \\htmlData{anchor=diff-2-rhs}{\\color{@amber}{(4p + 3) - (3p - 1)}}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"diff-1-lhs"},{"color":"amber","style":"box","anchor":"diff-2-rhs"}],"connections":[{"to":"diff-2-rhs","from":"diff-1-lhs","color":"sky","label":"equal difference","style":"solid"}]},"order":1,"title":"Use common differences","description":"In an arithmetic sequence, the difference $d$ between consecutive terms is constant. We can set up an equation by equating the difference between the second and first terms to the difference between the third and second terms."},{"type":"text","order":2,"title":"Simplify both sides","content":"Now, let's simplify each side of our equation:\n\n* Left-hand side (LHS):\n $$(3p - 1) - (p + 5) = 3p - 1 - p - 5 = 2p - 6$$\n* Right-hand side (RHS):\n $$(4p + 3) - (3p - 1) = 4p + 3 - 3p + 1 = p + 4$$\n\nEquating these two simplified expressions gives:\n $$2p - 6 = p + 4$$"},{"type":"text","order":3,"title":"Solve for p","content":"To solve for $p$, subtract $p$ from both sides and add $6$ to both sides:\n\n$$\\begin{aligned} 2p - p &= 4 + 6 \\\\ p &= 10 \\end{aligned}$$"}]},{"id":"arithmetic_mean","label":"Using Arithmetic Mean Property","steps":[{"type":"text","order":0,"title":"Recall arithmetic mean property","content":"For any three consecutive terms $a, b,$ and $c$ in an arithmetic sequence, the middle term is the arithmetic mean of the first and third terms.\n\nThis gives us the relationship:\n\n$$b = \\frac{a + c}{2}$$\n\nAlternatively, we can write this as:\n\n$$2b = a + c$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$a = \\htmlData{anchor=term1}{\\color{@sky}{p + 5}}, \\quad b = \\htmlData{anchor=term2}{\\color{@amber}{3p - 1}}, \\quad c = \\htmlData{anchor=term3}{\\color{@purple}{4p + 3}}$"},{"latex":"$$2(\\htmlData{anchor=mid-sub}{\\color{@amber}{3p - 1}}) = (\\htmlData{anchor=left-sub}{\\color{@sky}{p + 5}}) + (\\htmlData{anchor=right-sub}{\\color{@purple}{4p + 3}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"term1"},{"color":"amber","style":"text-color","anchor":"term2"},{"color":"purple","style":"text-color","anchor":"term3"}],"connections":[{"to":"left-sub","from":"term1","color":"sky","label":"First term","style":"solid"},{"to":"mid-sub","from":"term2","color":"amber","label":"Middle term","style":"solid"},{"to":"right-sub","from":"term3","color":"purple","label":"Third term","style":"solid"}]},"order":1,"title":"Set up the equation","description":"We identify the three consecutive terms of the sequence and substitute them into the arithmetic mean relationship $2b = a + c$ to establish a linear equation in terms of $p$."},{"type":"text","order":2,"title":"Solve for p","content":"Now we simplify both sides of the equation and solve for $p$:\n\n$$\\begin{aligned} 2(3p - 1) &= (p + 5) + (4p + 3) \\\\ 6p - 2 &= 5p + 8 \\end{aligned}$$\n\nSubtract $5p$ from both sides:\n\n$$p - 2 = 8$$\n\nAdd $2$ to both sides to find our final answer:\n\n$$p = 10$$"}]},{"id":"system_substitution","label":"Parameter Substitution","steps":[{"type":"text","order":0,"title":"Define key parameters","content":"To use the parameter substitution approach, we begin by defining the key components of an arithmetic sequence:\n- The first term, $a$\n- The common difference, $d$\n\nBased on the given terms:\n- First term: $u_1 = p + 5$\n- Second term: $u_2 = 3p - 1$\n- Third term: $u_3 = 4p + 3$\n\nWe set our first term parameter as:\n$$a = p + 5$$"},{"type":"text","order":1,"title":"Find common difference","content":"The common difference $d$ is the difference between any two consecutive terms. We can define $d$ using the first two terms:\n\n$$d = u_2 - u_1$$\n\nSubstitute the given expressions for $u_1$ and $u_2$:\n\n$$\\begin{aligned} d &= (3p - 1) - (p + 5) \\\\ d &= 3p - 1 - p - 5 \\\\ d &= 2p - 6 \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Parameters: } \\htmlData{anchor=a_def}{\\color{@sky}{a = p + 5}}, \\quad \\htmlData{anchor=d_def}{\\color{@amber}{d = 2p - 6}}$"},{"latex":"$$u_3 = \\htmlData{anchor=a_sub}{\\color{@sky}{a}} + 2\\htmlData{anchor=d_sub}{\\color{@amber}{d}}$"},{"latex":"$$4p + 3 = \\htmlData{anchor=a_expr}{\\color{@sky}{(p + 5)}} + 2\\htmlData{anchor=d_expr}{\\color{@amber}{(2p - 6)}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a_def"},{"color":"sky","style":"text-color","anchor":"a_sub"},{"color":"sky","style":"text-color","anchor":"a_expr"},{"color":"amber","style":"text-color","anchor":"d_def"},{"color":"amber","style":"text-color","anchor":"d_sub"},{"color":"amber","style":"text-color","anchor":"d_expr"}],"connections":[{"to":"a_sub","from":"a_def","color":"sky","label":"substitute a","style":"solid"},{"to":"d_sub","from":"d_def","color":"amber","label":"substitute d","style":"solid"}]},"order":2,"title":"Set up third term","description":"The general formula for the third term is $u_3 = a + 2d$. We substitute our expressions for $a$ and $d$ into this formula."},{"type":"text","order":3,"title":"Solve for p","content":"Now, we expand and simplify the equation to find the value of $p$:\n\n$$\\begin{aligned} 4p + 3 &= (p + 5) + 2(2p - 6) \\\\ 4p + 3 &= p + 5 + 4p - 12 \\\\ 4p + 3 &= 5p - 7 \\end{aligned}$$\n\nRearrange to solve for $p$:\n\n$$\\begin{aligned} 3 + 7 &= 5p - 4p \\\\ 10 &= p \\end{aligned}$$\n\nThus, $p = 10$."}]}],"generatedAt":"2026-06-09T01:57:40.772Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"437b0296-984b-4e68-beb6-66377a34f47b","label":"b","content":"Find the sum of the first 12 terms of the series.","markscheme":"- Find $a = 10+5 = 15$ and $d = (3(10)-1) - 15 = 14$ M1\n- $S_{12} = \\frac{12}{2}(2(15) + 11(14))$ M1\n- $S_{12} = 6(30 + 154) = 1104$ A1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p3","marks":1,"description":"Finds the first term $a$ and the common difference $d$ using their value of $p$.","code":"M"},{"id":"p4","marks":1,"description":"Correctly substitutes $a$, $d$, and $n=12$ into the arithmetic sum formula.","code":"M"},{"id":"p5","marks":1,"description":"Calculates the correct final sum of 1104.","code":"A"}],"finalAnswers":["1104"],"generatedAt":"2026-05-06T14:05:22.840Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"standard_sum_formula","label":"Using standard sum formula","steps":[{"type":"text","order":0,"title":"Identify sequence parameters","content":"From the previous part of the question, we found $p = 10$.\n\nLet's first determine the first term, $a$, and the common difference, $d$, of our arithmetic series:\n- **First term ($a$)**:\n $$a = p + 5 = 10 + 5 = 15$$\n- **Second term**:\n $$3p - 1 = 3(10) - 1 = 29$$\n- **Common difference ($d$)**:\n $$d = 29 - 15 = 14$$\n\nWe want to find the sum of the first $n = 12$ terms."},{"type":"text","order":1,"title":"Recall the sum formula","content":"The sum of the first $n$ terms of an arithmetic series is given by the formula:\n\n$$S_n = \\frac{n}{2}\\big(2a + (n - 1)d\\big)$$\n\nwhere:\n- $a$ is the first term\n- $d$ is the common difference\n- $n$ is the number of terms"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=a-val}{\\color{@sky}{a = 15}}, \\quad \\htmlData{anchor=d-val}{\\color{@amber}{d = 14}}, \\quad \\htmlData{anchor=n-val}{\\color{@purple}{n = 12}}$"},{"latex":"$$S_{\\htmlData{anchor=n-sub1}{\\color{@purple}{12}}} = \\frac{\\htmlData{anchor=n-sub2}{\\color{@purple}{12}}}{2}\\Big(2(\\htmlData{anchor=a-sub}{\\color{@sky}{15}}) + (\\htmlData{anchor=n-sub3}{\\color{@purple}{12}} - 1)\\htmlData{anchor=d-sub}{\\color{@amber}{14}}\\Big)$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"amber","style":"text-color","anchor":"d-val"},{"color":"purple","style":"text-color","anchor":"n-val"}],"connections":[{"to":"a-sub","from":"a-val","color":"sky","label":"Substitute a","style":"solid"},{"to":"d-sub","from":"d-val","color":"amber","label":"Substitute d","style":"solid"},{"to":"n-sub1","from":"n-val","color":"purple","label":"Substitute n","style":"solid"},{"to":"n-sub2","from":"n-val","color":"purple","label":null,"style":"solid"},{"to":"n-sub3","from":"n-val","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Substitute the values","description":"Substitute $a = 15$, $d = 14$, and $n = 12$ directly into the arithmetic series sum formula."},{"type":"text","order":3,"title":"Calculate the final sum","content":"Now, we simplify the expression step-by-step:\n\n$$\\begin{aligned} S_{12} &= 6 \\big(2(15) + 11(14)\\big) \\\\ S_{12} &= 6(30 + 154) \\\\ S_{12} &= 6(184) \\\\ S_{12} &= 1104 \\end{aligned}$$\n\nThus, the sum of the first 12 terms is $1104$.","calculator":[{"gdc":{"expression":"12/2*(2*15 + 11*14)","instructions":"In the Run-Matrix menu, enter the calculation:\nType '12/2 * (2 * 15 + 11 * 14)' and press [EXE]."},"ti84":{"expression":"12/2*(2*15 + 11*14)","instructions":"Type the expression directly on the main screen:\nPress [1][2] [/] [2] [*] [(] [2] [*] [1][5] [+] [1][1] [*] [1][4] [)] and press [ENTER]."},"desmos":{"expression":"\\frac{12}{2}(2(15) + 11(14))","instructions":"In a blank expression bar, enter the calculation directly to see the result on the right side."},"description":"Perform the arithmetic calculation directly using a calculator."}]}]},{"id":"last_term_formula","label":"Finding the 12th term first","steps":[{"type":"text","order":0,"title":"Identify first term and difference","content":"From the previous part of the question, we found $p = 10$. Let's use this value to find the first term $a$ and the common difference $d$ of the arithmetic series:\n\n* First term, $$a = p + 5 = 10 + 5 = 15$$\n* Second term, $$u_2 = 3p - 1 = 3(10) - 1 = 29$$\n\nNow, we calculate the common difference $d$:\n\n$$d = u_2 - a = 29 - 15 = 14$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$a = \\htmlData{anchor=a-val}{\\color{@sky}{15}}, \\quad d = \\htmlData{anchor=d-val}{\\color{@amber}{14}}, \\quad n = \\htmlData{anchor=n-val}{\\color{@purple}{12}}$"},{"latex":"$$u_{12} = \\htmlData{anchor=a-formula}{\\color{@sky}{a}} + (n - 1)\\htmlData{anchor=d-formula}{\\color{@amber}{d}}$"},{"latex":"$$u_{12} = \\color{@sky}{15} + (\\color{@purple}{12} - 1)(\\color{@amber}{14})$"},{"latex":"$$u_{12} = 15 + 11(14) = 15 + 154 = 169$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"amber","style":"text-color","anchor":"d-val"},{"color":"purple","style":"text-color","anchor":"n-val"}],"connections":[{"to":"a-formula","from":"a-val","color":"sky","label":null,"style":"solid"},{"to":"d-formula","from":"d-val","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Calculate the 12th term","description":"Before finding the sum, we calculate the $12^{\\text{th}}$ term ($u_{12}$) using the general $n^{\\text{th}}$ term formula: $$u_n = a + (n-1)d$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$a = \\htmlData{anchor=a-val2}{\\color{@sky}{15}}, \\quad u_{12} = \\htmlData{anchor=u-val2}{\\color{@teal}{169}}, \\quad n = \\htmlData{anchor=n-val2}{\\color{@purple}{12}}$"},{"latex":"$$S_{12} = \\frac{\\htmlData{anchor=n-formula2}{\\color{@purple}{n}}}{2}\\left(\\htmlData{anchor=a-formula2}{\\color{@sky}{a}} + \\htmlData{anchor=u-formula2}{\\color{@teal}{u_{12}}}\\right)$"},{"latex":"$$S_{12} = \\frac{\\color{@purple}{12}}{2}(\\color{@sky}{15} + \\color{@teal}{169})$"},{"latex":"$$S_{12} = 6(184) = 1104$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val2"},{"color":"teal","style":"text-color","anchor":"u-val2"},{"color":"purple","style":"text-color","anchor":"n-val2"}],"connections":[{"to":"a-formula2","from":"a-val2","color":"sky","label":null,"style":"solid"},{"to":"u-formula2","from":"u-val2","color":"teal","label":null,"style":"solid"},{"to":"n-formula2","from":"n-val2","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Calculate the sum","description":"Now, we use the alternative sum formula to find $S_{12}$: $$S_n = \\frac{n}{2}(a + u_n)$$"}]},{"id":"gdc_summation","label":"Using summation on GDC","steps":[{"type":"text","order":0,"title":"Recall previous values","content":"From the first part of the question, we found the value of $p$ is $10$.\n\nLet's first calculate the first term ($a$) and the common difference ($d$) of our arithmetic series using $p = 10$:\n\n* First term, $a = p + 5 = 10 + 5 = 15$\n* Second term, $u_2 = 3p - 1 = 3(10) - 1 = 29$\n* Common difference, $d = 29 - 15 = 14$\n\nThis gives us an arithmetic series with a first term of $15$ and a common difference of $14$."},{"type":"text","order":1,"title":"Formulate general term","content":"The general term $u_r$ of an arithmetic sequence is given by:\n\n$$u_r = a + (r - 1)d$$\n\nSubstituting our known values, $a = 15$ and $d = 14$:\n\n$$\\begin{aligned} u_r &= 15 + (r - 1)14 \\\\ &= 15 + 14r - 14 \\\\ &= 14r + 1 \\end{aligned}$$\n\nSo, the general term is $u_r = 14r + 1$."},{"type":"text","order":2,"title":"Set up sum notation","content":"We want to find the sum of the first $12$ terms. \n\nWe can write this sum using sigma notation as:\n\n$$S_{12} = \\sum_{r=1}^{12} u_r = \\sum_{r=1}^{12} (14r + 1)$$"},{"type":"text","order":3,"title":"Evaluate using GDC","content":"Now we can enter this summation directly into our graphing display calculator (GDC) to find the answer swiftly.","calculator":[{"gdc":{"expression":"\\sum_{X=1}^{12} (14X + 1)","instructions":"1. Go to the Run-Matrix menu.\n2. Press [MATH] (F4), then [▷] (F6) to see more options, and select [Σ] (F2).\n3. Fill in the template: variable as X, lower bound as 1, upper bound as 12, and formula as 14X + 1.\n4. Press [EXE]."},"ti84":{"expression":"\\sum_{X=1}^{12} (14X + 1)","instructions":"1. Press [ALPHA] then [WINDOW] (F2).\n2. Select 2: summation Σ(\n3. Input the expression: Σ(X, 1, 12, 14X + 1)\n4. Press [ENTER]."},"desmos":{"expression":"\\sum_{n=1}^{12}(14n+1)","instructions":"Type 'sum' to bring up the summation notation, then fill in the limits from n=1 to 12 with the expression 14n + 1."},"description":"Calculate the sum of the series using the summation template."}]},{"type":"text","order":4,"title":"State final answer","content":"The GDC outputs:\n\n$$S_{12} = 1104$$"}]},{"id":"manual_addition","label":"Listing and adding terms","steps":[{"type":"text","order":0,"title":"Identify first term and difference","content":"From the first part of the question, we know:\n- The first term ($a$) is $15$.\n- The common difference ($d$) is $14$.\n\nTo find the sum of the first $12$ terms by listing them, we will start with our first term and repeatedly add $14$ to find each subsequent term until we have all $12$ terms."},{"type":"text","order":1,"title":"List all 12 terms","content":"Let's list the first $12$ terms by adding $14$ step-by-step:\n\n- Term 1: $15$\n- Term 2: $15 + 14 = 29$\n- Term 3: $29 + 14 = 43$\n- Term 4: $43 + 14 = 57$\n- Term 5: $57 + 14 = 71$\n- Term 6: $71 + 14 = 85$\n- Term 7: $85 + 14 = 99$\n- Term 8: $99 + 14 = 113$\n- Term 9: $113 + 14 = 127$\n- Term 10: $127 + 14 = 141$\n- Term 11: $141 + 14 = 155$\n- Term 12: $155 + 14 = 169$\n\nSo, our $12$ terms are:\n$$15, 29, 43, 57, 71, 85, 99, 113, 127, 141, 155, 169$$"},{"type":"text","order":2,"title":"Sum the listed terms","content":"Now, we find the sum of these $12$ terms by adding them together:\n\n$$\\begin{aligned} S_{12} &= 15 + 29 + 43 + 57 + 71 + 85 + 99 + 113 + 127 + 141 + 155 + 169 \\\\ S_{12} &= 1104 \\end{aligned}$$\n\nThis gives us our final sum of $1104$.","calculator":[{"gdc":{"expression":"15+29+43+57+71+85+99+113+127+141+155+169","instructions":"Enter the sum of the list directly on the main calculation screen:\n15 + 29 + 43 + 57 + 71 + 85 + 99 + 113 + 127 + 141 + 155 + 169"},"ti84":{"expression":"15+29+43+57+71+85+99+113+127+141+155+169","instructions":"Type the numbers directly into the home screen and add them:\n15 + 29 + 43 + 57 + 71 + 85 + 99 + 113 + 127 + 141 + 155 + 169 [ENTER]"},"desmos":{"expression":"15+29+43+57+71+85+99+113+127+141+155+169","instructions":"Type the addition expression in a single line to calculate the total sum."},"description":"You can quickly sum this list of numbers using your calculator."}]}]}],"generatedAt":"2026-06-09T08:48:53.369Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"739eff19-09b7-4564-bf50-3192c3eff459:common_difference","label":"(a) Equating Common Differences","steps":[{"type":"text","order":0,"title":"Identify the terms","content":"The question gives us the first three terms of an arithmetic series:\n\n* First term: $u_1 = p + 5$\n* Second term: $u_2 = 3p - 1$\n* Third term: $u_3 = 4p + 3$","highlights":[{"text":"(p+5)","location":"specification"},{"text":"(3p-1)","location":"specification"},{"text":"(4p+3)","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$d = \\text{Term } 2 - \\text{Term } 1 = \\text{Term } 3 - \\text{Term } 2$$"},{"latex":"$$$\\htmlData{anchor=diff-1-lhs}{\\color{@sky}{(3p - 1) - (p + 5)}} = \\htmlData{anchor=diff-2-rhs}{\\color{@amber}{(4p + 3) - (3p - 1)}}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"diff-1-lhs"},{"color":"amber","style":"box","anchor":"diff-2-rhs"}],"connections":[{"to":"diff-2-rhs","from":"diff-1-lhs","color":"sky","label":"equal difference","style":"solid"}]},"order":1,"title":"Use common differences","description":"In an arithmetic sequence, the difference $d$ between consecutive terms is constant. We can set up an equation by equating the difference between the second and first terms to the difference between the third and second terms."},{"type":"text","order":2,"title":"Simplify both sides","content":"Now, let's simplify each side of our equation:\n\n* Left-hand side (LHS):\n $$(3p - 1) - (p + 5) = 3p - 1 - p - 5 = 2p - 6$$\n* Right-hand side (RHS):\n $$(4p + 3) - (3p - 1) = 4p + 3 - 3p + 1 = p + 4$$\n\nEquating these two simplified expressions gives:\n $$2p - 6 = p + 4$$"},{"type":"text","order":3,"title":"Solve for p","content":"To solve for $p$, subtract $p$ from both sides and add $6$ to both sides:\n\n$$\\begin{aligned} 2p - p &= 4 + 6 \\\\ p &= 10 \\end{aligned}$$"}]},{"id":"739eff19-09b7-4564-bf50-3192c3eff459:arithmetic_mean","label":"(a) Using Arithmetic Mean Property","steps":[{"type":"text","order":0,"title":"Recall arithmetic mean property","content":"For any three consecutive terms $a, b,$ and $c$ in an arithmetic sequence, the middle term is the arithmetic mean of the first and third terms.\n\nThis gives us the relationship:\n\n$$b = \\frac{a + c}{2}$$\n\nAlternatively, we can write this as:\n\n$$2b = a + c$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$a = \\htmlData{anchor=term1}{\\color{@sky}{p + 5}}, \\quad b = \\htmlData{anchor=term2}{\\color{@amber}{3p - 1}}, \\quad c = \\htmlData{anchor=term3}{\\color{@purple}{4p + 3}}$"},{"latex":"$$2(\\htmlData{anchor=mid-sub}{\\color{@amber}{3p - 1}}) = (\\htmlData{anchor=left-sub}{\\color{@sky}{p + 5}}) + (\\htmlData{anchor=right-sub}{\\color{@purple}{4p + 3}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"term1"},{"color":"amber","style":"text-color","anchor":"term2"},{"color":"purple","style":"text-color","anchor":"term3"}],"connections":[{"to":"left-sub","from":"term1","color":"sky","label":"First term","style":"solid"},{"to":"mid-sub","from":"term2","color":"amber","label":"Middle term","style":"solid"},{"to":"right-sub","from":"term3","color":"purple","label":"Third term","style":"solid"}]},"order":1,"title":"Set up the equation","description":"We identify the three consecutive terms of the sequence and substitute them into the arithmetic mean relationship $2b = a + c$ to establish a linear equation in terms of $p$."},{"type":"text","order":2,"title":"Solve for p","content":"Now we simplify both sides of the equation and solve for $p$:\n\n$$\\begin{aligned} 2(3p - 1) &= (p + 5) + (4p + 3) \\\\ 6p - 2 &= 5p + 8 \\end{aligned}$$\n\nSubtract $5p$ from both sides:\n\n$$p - 2 = 8$$\n\nAdd $2$ to both sides to find our final answer:\n\n$$p = 10$$"}]},{"id":"739eff19-09b7-4564-bf50-3192c3eff459:system_substitution","label":"(a) Parameter Substitution","steps":[{"type":"text","order":0,"title":"Define key parameters","content":"To use the parameter substitution approach, we begin by defining the key components of an arithmetic sequence:\n- The first term, $a$\n- The common difference, $d$\n\nBased on the given terms:\n- First term: $u_1 = p + 5$\n- Second term: $u_2 = 3p - 1$\n- Third term: $u_3 = 4p + 3$\n\nWe set our first term parameter as:\n$$a = p + 5$$"},{"type":"text","order":1,"title":"Find common difference","content":"The common difference $d$ is the difference between any two consecutive terms. We can define $d$ using the first two terms:\n\n$$d = u_2 - u_1$$\n\nSubstitute the given expressions for $u_1$ and $u_2$:\n\n$$\\begin{aligned} d &= (3p - 1) - (p + 5) \\\\ d &= 3p - 1 - p - 5 \\\\ d &= 2p - 6 \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Parameters: } \\htmlData{anchor=a_def}{\\color{@sky}{a = p + 5}}, \\quad \\htmlData{anchor=d_def}{\\color{@amber}{d = 2p - 6}}$"},{"latex":"$$u_3 = \\htmlData{anchor=a_sub}{\\color{@sky}{a}} + 2\\htmlData{anchor=d_sub}{\\color{@amber}{d}}$"},{"latex":"$$4p + 3 = \\htmlData{anchor=a_expr}{\\color{@sky}{(p + 5)}} + 2\\htmlData{anchor=d_expr}{\\color{@amber}{(2p - 6)}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a_def"},{"color":"sky","style":"text-color","anchor":"a_sub"},{"color":"sky","style":"text-color","anchor":"a_expr"},{"color":"amber","style":"text-color","anchor":"d_def"},{"color":"amber","style":"text-color","anchor":"d_sub"},{"color":"amber","style":"text-color","anchor":"d_expr"}],"connections":[{"to":"a_sub","from":"a_def","color":"sky","label":"substitute a","style":"solid"},{"to":"d_sub","from":"d_def","color":"amber","label":"substitute d","style":"solid"}]},"order":2,"title":"Set up third term","description":"The general formula for the third term is $u_3 = a + 2d$. We substitute our expressions for $a$ and $d$ into this formula."},{"type":"text","order":3,"title":"Solve for p","content":"Now, we expand and simplify the equation to find the value of $p$:\n\n$$\\begin{aligned} 4p + 3 &= (p + 5) + 2(2p - 6) \\\\ 4p + 3 &= p + 5 + 4p - 12 \\\\ 4p + 3 &= 5p - 7 \\end{aligned}$$\n\nRearrange to solve for $p$:\n\n$$\\begin{aligned} 3 + 7 &= 5p - 4p \\\\ 10 &= p \\end{aligned}$$\n\nThus, $p = 10$."}]},{"id":"437b0296-984b-4e68-beb6-66377a34f47b:standard_sum_formula","label":"(b) Using standard sum formula","steps":[{"type":"text","order":0,"title":"Identify sequence parameters","content":"From the previous part of the question, we found $p = 10$.\n\nLet's first determine the first term, $a$, and the common difference, $d$, of our arithmetic series:\n- **First term ($a$)**:\n $$a = p + 5 = 10 + 5 = 15$$\n- **Second term**:\n $$3p - 1 = 3(10) - 1 = 29$$\n- **Common difference ($d$)**:\n $$d = 29 - 15 = 14$$\n\nWe want to find the sum of the first $n = 12$ terms."},{"type":"text","order":1,"title":"Recall the sum formula","content":"The sum of the first $n$ terms of an arithmetic series is given by the formula:\n\n$$S_n = \\frac{n}{2}\\big(2a + (n - 1)d\\big)$$\n\nwhere:\n- $a$ is the first term\n- $d$ is the common difference\n- $n$ is the number of terms"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=a-val}{\\color{@sky}{a = 15}}, \\quad \\htmlData{anchor=d-val}{\\color{@amber}{d = 14}}, \\quad \\htmlData{anchor=n-val}{\\color{@purple}{n = 12}}$"},{"latex":"$$S_{\\htmlData{anchor=n-sub1}{\\color{@purple}{12}}} = \\frac{\\htmlData{anchor=n-sub2}{\\color{@purple}{12}}}{2}\\Big(2(\\htmlData{anchor=a-sub}{\\color{@sky}{15}}) + (\\htmlData{anchor=n-sub3}{\\color{@purple}{12}} - 1)\\htmlData{anchor=d-sub}{\\color{@amber}{14}}\\Big)$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"amber","style":"text-color","anchor":"d-val"},{"color":"purple","style":"text-color","anchor":"n-val"}],"connections":[{"to":"a-sub","from":"a-val","color":"sky","label":"Substitute a","style":"solid"},{"to":"d-sub","from":"d-val","color":"amber","label":"Substitute d","style":"solid"},{"to":"n-sub1","from":"n-val","color":"purple","label":"Substitute n","style":"solid"},{"to":"n-sub2","from":"n-val","color":"purple","label":null,"style":"solid"},{"to":"n-sub3","from":"n-val","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Substitute the values","description":"Substitute $a = 15$, $d = 14$, and $n = 12$ directly into the arithmetic series sum formula."},{"type":"text","order":3,"title":"Calculate the final sum","content":"Now, we simplify the expression step-by-step:\n\n$$\\begin{aligned} S_{12} &= 6 \\big(2(15) + 11(14)\\big) \\\\ S_{12} &= 6(30 + 154) \\\\ S_{12} &= 6(184) \\\\ S_{12} &= 1104 \\end{aligned}$$\n\nThus, the sum of the first 12 terms is $1104$.","calculator":[{"gdc":{"expression":"12/2*(2*15 + 11*14)","instructions":"In the Run-Matrix menu, enter the calculation:\nType '12/2 * (2 * 15 + 11 * 14)' and press [EXE]."},"ti84":{"expression":"12/2*(2*15 + 11*14)","instructions":"Type the expression directly on the main screen:\nPress [1][2] [/] [2] [*] [(] [2] [*] [1][5] [+] [1][1] [*] [1][4] [)] and press [ENTER]."},"desmos":{"expression":"\\frac{12}{2}(2(15) + 11(14))","instructions":"In a blank expression bar, enter the calculation directly to see the result on the right side."},"description":"Perform the arithmetic calculation directly using a calculator."}]}]},{"id":"437b0296-984b-4e68-beb6-66377a34f47b:last_term_formula","label":"(b) Finding the 12th term first","steps":[{"type":"text","order":0,"title":"Identify first term and difference","content":"From the previous part of the question, we found $p = 10$. Let's use this value to find the first term $a$ and the common difference $d$ of the arithmetic series:\n\n* First term, $$a = p + 5 = 10 + 5 = 15$$\n* Second term, $$u_2 = 3p - 1 = 3(10) - 1 = 29$$\n\nNow, we calculate the common difference $d$:\n\n$$d = u_2 - a = 29 - 15 = 14$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$a = \\htmlData{anchor=a-val}{\\color{@sky}{15}}, \\quad d = \\htmlData{anchor=d-val}{\\color{@amber}{14}}, \\quad n = \\htmlData{anchor=n-val}{\\color{@purple}{12}}$"},{"latex":"$$u_{12} = \\htmlData{anchor=a-formula}{\\color{@sky}{a}} + (n - 1)\\htmlData{anchor=d-formula}{\\color{@amber}{d}}$"},{"latex":"$$u_{12} = \\color{@sky}{15} + (\\color{@purple}{12} - 1)(\\color{@amber}{14})$"},{"latex":"$$u_{12} = 15 + 11(14) = 15 + 154 = 169$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"amber","style":"text-color","anchor":"d-val"},{"color":"purple","style":"text-color","anchor":"n-val"}],"connections":[{"to":"a-formula","from":"a-val","color":"sky","label":null,"style":"solid"},{"to":"d-formula","from":"d-val","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Calculate the 12th term","description":"Before finding the sum, we calculate the $12^{\\text{th}}$ term ($u_{12}$) using the general $n^{\\text{th}}$ term formula: $$u_n = a + (n-1)d$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$a = \\htmlData{anchor=a-val2}{\\color{@sky}{15}}, \\quad u_{12} = \\htmlData{anchor=u-val2}{\\color{@teal}{169}}, \\quad n = \\htmlData{anchor=n-val2}{\\color{@purple}{12}}$"},{"latex":"$$S_{12} = \\frac{\\htmlData{anchor=n-formula2}{\\color{@purple}{n}}}{2}\\left(\\htmlData{anchor=a-formula2}{\\color{@sky}{a}} + \\htmlData{anchor=u-formula2}{\\color{@teal}{u_{12}}}\\right)$"},{"latex":"$$S_{12} = \\frac{\\color{@purple}{12}}{2}(\\color{@sky}{15} + \\color{@teal}{169})$"},{"latex":"$$S_{12} = 6(184) = 1104$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val2"},{"color":"teal","style":"text-color","anchor":"u-val2"},{"color":"purple","style":"text-color","anchor":"n-val2"}],"connections":[{"to":"a-formula2","from":"a-val2","color":"sky","label":null,"style":"solid"},{"to":"u-formula2","from":"u-val2","color":"teal","label":null,"style":"solid"},{"to":"n-formula2","from":"n-val2","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Calculate the sum","description":"Now, we use the alternative sum formula to find $S_{12}$: $$S_n = \\frac{n}{2}(a + u_n)$$"}]},{"id":"437b0296-984b-4e68-beb6-66377a34f47b:gdc_summation","label":"(b) Using summation on GDC","steps":[{"type":"text","order":0,"title":"Recall previous values","content":"From the first part of the question, we found the value of $p$ is $10$.\n\nLet's first calculate the first term ($a$) and the common difference ($d$) of our arithmetic series using $p = 10$:\n\n* First term, $a = p + 5 = 10 + 5 = 15$\n* Second term, $u_2 = 3p - 1 = 3(10) - 1 = 29$\n* Common difference, $d = 29 - 15 = 14$\n\nThis gives us an arithmetic series with a first term of $15$ and a common difference of $14$."},{"type":"text","order":1,"title":"Formulate general term","content":"The general term $u_r$ of an arithmetic sequence is given by:\n\n$$u_r = a + (r - 1)d$$\n\nSubstituting our known values, $a = 15$ and $d = 14$:\n\n$$\\begin{aligned} u_r &= 15 + (r - 1)14 \\\\ &= 15 + 14r - 14 \\\\ &= 14r + 1 \\end{aligned}$$\n\nSo, the general term is $u_r = 14r + 1$."},{"type":"text","order":2,"title":"Set up sum notation","content":"We want to find the sum of the first $12$ terms. \n\nWe can write this sum using sigma notation as:\n\n$$S_{12} = \\sum_{r=1}^{12} u_r = \\sum_{r=1}^{12} (14r + 1)$$"},{"type":"text","order":3,"title":"Evaluate using GDC","content":"Now we can enter this summation directly into our graphing display calculator (GDC) to find the answer swiftly.","calculator":[{"gdc":{"expression":"\\sum_{X=1}^{12} (14X + 1)","instructions":"1. Go to the Run-Matrix menu.\n2. Press [MATH] (F4), then [▷] (F6) to see more options, and select [Σ] (F2).\n3. Fill in the template: variable as X, lower bound as 1, upper bound as 12, and formula as 14X + 1.\n4. Press [EXE]."},"ti84":{"expression":"\\sum_{X=1}^{12} (14X + 1)","instructions":"1. Press [ALPHA] then [WINDOW] (F2).\n2. Select 2: summation Σ(\n3. Input the expression: Σ(X, 1, 12, 14X + 1)\n4. Press [ENTER]."},"desmos":{"expression":"\\sum_{n=1}^{12}(14n+1)","instructions":"Type 'sum' to bring up the summation notation, then fill in the limits from n=1 to 12 with the expression 14n + 1."},"description":"Calculate the sum of the series using the summation template."}]},{"type":"text","order":4,"title":"State final answer","content":"The GDC outputs:\n\n$$S_{12} = 1104$$"}]},{"id":"437b0296-984b-4e68-beb6-66377a34f47b:manual_addition","label":"(b) Listing and adding terms","steps":[{"type":"text","order":0,"title":"Identify first term and difference","content":"From the first part of the question, we know:\n- The first term ($a$) is $15$.\n- The common difference ($d$) is $14$.\n\nTo find the sum of the first $12$ terms by listing them, we will start with our first term and repeatedly add $14$ to find each subsequent term until we have all $12$ terms."},{"type":"text","order":1,"title":"List all 12 terms","content":"Let's list the first $12$ terms by adding $14$ step-by-step:\n\n- Term 1: $15$\n- Term 2: $15 + 14 = 29$\n- Term 3: $29 + 14 = 43$\n- Term 4: $43 + 14 = 57$\n- Term 5: $57 + 14 = 71$\n- Term 6: $71 + 14 = 85$\n- Term 7: $85 + 14 = 99$\n- Term 8: $99 + 14 = 113$\n- Term 9: $113 + 14 = 127$\n- Term 10: $127 + 14 = 141$\n- Term 11: $141 + 14 = 155$\n- Term 12: $155 + 14 = 169$\n\nSo, our $12$ terms are:\n$$15, 29, 43, 57, 71, 85, 99, 113, 127, 141, 155, 169$$"},{"type":"text","order":2,"title":"Sum the listed terms","content":"Now, we find the sum of these $12$ terms by adding them together:\n\n$$\\begin{aligned} S_{12} &= 15 + 29 + 43 + 57 + 71 + 85 + 99 + 113 + 127 + 141 + 155 + 169 \\\\ S_{12} &= 1104 \\end{aligned}$$\n\nThis gives us our final sum of $1104$.","calculator":[{"gdc":{"expression":"15+29+43+57+71+85+99+113+127+141+155+169","instructions":"Enter the sum of the list directly on the main calculation screen:\n15 + 29 + 43 + 57 + 71 + 85 + 99 + 113 + 127 + 141 + 155 + 169"},"ti84":{"expression":"15+29+43+57+71+85+99+113+127+141+155+169","instructions":"Type the numbers directly into the home screen and add them:\n15 + 29 + 43 + 57 + 71 + 85 + 99 + 113 + 127 + 141 + 155 + 169 [ENTER]"},"desmos":{"expression":"15+29+43+57+71+85+99+113+127+141+155+169","instructions":"Type the addition expression in a single line to calculate the total sum."},"description":"You can quickly sum this list of numbers using your calculator."}]}]}],"generatedAt":"2026-06-09T08:49:03.482Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"b1c78b39-7ff3-4d13-a0c8-6383e71a97bd","revisionId":"8e6a64ab-d349-4a67-a7bc-f436ba30b656","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"547ecc9b-6a40-45eb-b47f-0ca1f4a971bc","label":"a","content":"Expand $(1+4x)^{-2}$ in ascending powers of $x$ up to and including $x^3$","markscheme":"- Applies expansion formula: $1 + (-2)(4x) + \\frac{(-2)(-3)}{2}(4x)^2 + \\frac{(-2)(-3)(-4)}{6}(4x)^3$ M1\n- Correct $x$ term: $-8x$ A1\n- Correct $x^2$ term: $48x^2$ A1\n- Correct $x^3$ term: $-256x^3$ A1","marks":4,"order":0,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Expansion formula used with n=-2 and (4x).","code":"M"},{"id":"p2","marks":1,"description":"Correct x term: -8x.","code":"A"},{"id":"p3","marks":1,"description":"Correct x^2 term: 48x^2.","code":"A"},{"id":"p4","marks":1,"description":"Correct x^3 term: -256x^3.","code":"A"}],"finalAnswers":["1-8x+48x^2-256x^3"],"generatedAt":"2026-05-06T14:04:03.147Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"binomial_expansion","label":"Binomial Series Expansion","steps":[{"type":"text","order":0,"title":"State the formula","content":"To expand $(1+4x)^{-2}$, we use the general binomial series expansion for a rational or negative index $n$:\n\n$$(1+u)^n = 1 + nu + \\frac{n(n-1)}{2!}u^2 + \\frac{n(n-1)(n-2)}{3!}u^3 + \\dots$$\n\nThis formula is valid when $|u| < 1$. Here, we will substitute $u = 4x$ and $n = -2$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u = \\htmlData{anchor=u-val}{\\color{@sky}{4x}}, \\quad n = \\htmlData{anchor=n-val}{\\color{@amber}{-2}}$"},{"latex":"$$(1 + \\htmlData{anchor=u-sub}{\\color{@sky}{4x}})^{\\htmlData{anchor=n-sub}{\\color{@amber}{-2}}} = 1 + (\\htmlData{anchor=n-term1}{\\color{@amber}{-2}})(\\htmlData{anchor=u-term1}{\\color{@sky}{4x}}) + \\frac{(\\htmlData{anchor=n-term2a}{\\color{@amber}{-2}})(\\htmlData{anchor=n-term2b}{\\color{@amber}{-3}})}{2}(\\htmlData{anchor=u-term2}{\\color{@sky}{4x}})^2 + \\frac{(\\htmlData{anchor=n-term3a}{\\color{@amber}{-2}})(\\htmlData{anchor=n-term3b}{\\color{@amber}{-3}})(\\htmlData{anchor=n-term3c}{\\color{@amber}{-4}})}{6}(\\htmlData{anchor=u-term3}{\\color{@sky}{4x}})^3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-val"},{"color":"amber","style":"text-color","anchor":"n-val"}],"connections":[{"to":"u-sub","from":"u-val","color":"sky","label":null,"style":"solid"},{"to":"n-sub","from":"n-val","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Substitute the values","description":"We substitute $u = 4x$ and $n = -2$ into the binomial expansion formula. Writing down this structured substitution earns the first method mark."},{"type":"text","order":2,"title":"Simplify each term","content":"Now we simplify each term step-by-step:\n\n* **$x$ term:**\n $$(-2)(4x) = -8x$$\n\n* **$x^2$ term:**\n $$\\frac{(-2)(-3)}{2}(4x)^2 = \\frac{6}{2}(16x^2) = 3(16x^2) = 48x^2$$\n\n* **$x^3$ term:**\n $$\\frac{(-2)(-3)(-4)}{6}(4x)^3 = \\frac{-24}{6}(64x^3) = -4(64x^3) = -256x^3$$"},{"type":"text","order":3,"title":"State final expansion","content":"Combining all the simplified terms, we get the final expansion up to the $x^3$ term:\n\n$$(1+4x)^{-2} = 1 - 8x + 48x^2 - 256x^3 + \\dots$$\n\nEach correct term earns an accuracy mark in the markscheme."}]},{"id":"maclaurin_series","label":"Maclaurin Series Expansion","steps":[{"type":"text","order":0,"title":"Define the Maclaurin Series","content":"To find the expansion of $f(x) = (1+4x)^{-2}$ up to the term in $x^3$ using the Maclaurin series, we use the standard formula:\n\n$$f(x) = f(0) + f'(0)x + \\frac{f''(0)}{2!}x^2 + \\frac{f'''(0)}{3!}x^3 + \\dots$$\n\nFirst, let's establish our function and evaluate it at $x = 0$:\n\n$$f(x) = (1+4x)^{-2}$$\n\n$$f(0) = (1+0)^{-2} = 1$$"},{"type":"text","order":1,"title":"Find the first derivative","content":"Using the chain rule, we differentiate $f(x) = (1+4x)^{-2}$:\n\n$$f'(x) = -2(1+4x)^{-3} \\cdot \\frac{d}{dx}(1+4x)$$\n\n$$f'(x) = -2(1+4x)^{-3} \\cdot 4 = -8(1+4x)^{-3}$$\n\nNow, evaluate $f'(x)$ at $x = 0$:\n\n$$f'(0) = -8(1+0)^{-3} = -8$$"},{"type":"text","order":2,"title":"Find the second derivative","content":"Next, we differentiate $f'(x) = -8(1+4x)^{-3}$ to find the second derivative:\n\n$$f''(x) = -8 \\cdot (-3)(1+4x)^{-4} \\cdot 4$$\n\n$$f''(x) = 24 \\cdot 4(1+4x)^{-4} = 96(1+4x)^{-4}$$\n\nNow, evaluate $f''(x)$ at $x = 0$:\n\n$$f''(0) = 96(1+0)^{-4} = 96$$"},{"type":"text","order":3,"title":"Find the third derivative","content":"We differentiate $f''(x) = 96(1+4x)^{-4}$ one more time to find the third derivative:\n\n$$f'''(x) = 96 \\cdot (-4)(1+4x)^{-5} \\cdot 4$$\n\n$$f'''(x) = -384 \\cdot 4(1+4x)^{-5} = -1536(1+4x)^{-5}$$\n\nNow, evaluate $f'''(x)$ at $x = 0$:\n\n$$f'''(0) = -1536(1+0)^{-5} = -1536$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(0) = \\htmlData{anchor=val0}{\\color{@sky}{1}}, \\quad f'(0) = \\htmlData{anchor=val1}{\\color{@amber}{-8}}, \\quad f''(0) = \\htmlData{anchor=val2}{\\color{@purple}{96}}, \\quad f'''(0) = \\htmlData{anchor=val3}{\\color{@teal}{-1536}}$"},{"latex":"$$f(x) \\approx \\htmlData{anchor=term0}{\\color{@sky}{1}} + \\htmlData{anchor=term1}{\\color{@amber}{(-8)}}x + \\frac{\\htmlData{anchor=term2}{\\color{@purple}{96}}}{2!}x^2 + \\frac{\\htmlData{anchor=term3}{\\color{@teal}{-1536}}}{3!}x^3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"val0"},{"color":"amber","style":"text-color","anchor":"val1"},{"color":"purple","style":"text-color","anchor":"val2"},{"color":"teal","style":"text-color","anchor":"val3"}],"connections":[{"to":"term0","from":"val0","color":"sky","label":null,"style":"solid"},{"to":"term1","from":"val1","color":"amber","label":null,"style":"solid"},{"to":"term2","from":"val2","color":"purple","label":null,"style":"solid"},{"to":"term3","from":"val3","color":"teal","label":null,"style":"solid"}]},"order":4,"title":"Substitute into Maclaurin Formula","description":"We substitute our calculated values for $f(0)$, $f'(0)$, $f''(0)$, and $f'''(0)$ directly into the Maclaurin series expansion formula."},{"type":"text","order":5,"title":"Simplify terms to final answer","content":"Now, simplify the coefficients to obtain the final expanded expression:\n\n* For the $x^2$ term: \n$$\\frac{96}{2!}x^2 = \\frac{96}{2}x^2 = 48x^2$$\n\n* For the $x^3$ term: \n$$\\frac{-1536}{3!}x^3 = \\frac{-1536}{6}x^3 = -256x^3$$\n\nPutting it all together, we get:\n\n$$(1+4x)^{-2} \\approx 1 - 8x + 48x^2 - 256x^3$$"}]},{"id":"long_division","label":"Algebraic Long Division","steps":[{"type":"text","order":0,"title":"Rewrite the expression","content":"To expand $(1+4x)^{-2}$ using algebraic division, we first rewrite the negative power as a fraction:\n\n$$(1+4x)^{-2} = \\frac{1}{(1+4x)^2}$$","highlights":[{"text":"$$(1+4x)^{-2}$","location":"specification"}]},{"type":"text","order":1,"title":"Expand the denominator","content":"Next, we expand the quadratic expression in the denominator:\n\n$$(1+4x)^2 = 1 + 2(4x) + (4x)^2 = 1 + 8x + 16x^2$$\n\nSubstituting this back into our fraction gives:\n\n$$\\frac{1}{1 + 8x + 16x^2}$$"},{"type":"text","order":2,"title":"Perform algebraic division","content":"To perform the division of $1$ by $1 + 8x + 16x^2$, we divide term-by-term:\n\n1. **First term:** $1 \\div 1 = 1$.\n * Subtracting $1(1 + 8x + 16x^2)$ from $1$ leaves: $$-8x - 16x^2$$\n\n2. **Second term:** $-8x \\div 1 = -8x$.\n * Subtracting $-8x(1 + 8x + 16x^2) = -8x - 64x^2 - 128x^3$ from $-8x - 16x^2$ leaves: $$48x^2 + 128x^3$$\n\n3. **Third term:** $48x^2 \\div 1 = 48x^2$.\n * Subtracting $48x^2(1 + 8x + 16x^2) = 48x^2 + 384x^3 + 768x^4$ from $48x^2 + 128x^3$ leaves: $$-256x^3 - 768x^4$$\n\n4. **Fourth term:** $-256x^3 \\div 1 = -256x^3$.\n\nCombining these quotients yields the series up to the $x^3$ term:\n$$1 - 8x + 48x^2 - 256x^3 + \\dots$$"},{"type":"text","order":3,"title":"State the final expansion","content":"From the quotient of our division, the expansion of $(1+4x)^{-2}$ up to the term in $x^3$ is:\n\n$$1 - 8x + 48x^2 - 256x^3$$"}]}],"generatedAt":"2026-06-09T08:09:14.104Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"3f556779-73ca-4be1-a8f0-b05439815f71","label":"b","content":"Write down the range of values of $x$ for which the expansion is valid.","markscheme":"- $|4x| < 1 \\Rightarrow |x| < \\frac{1}{4}$ B1","marks":1,"order":1,"markingRubric":{"totalMarks":1,"points":[{"id":"p1","marks":1,"description":"Correct range |x| < 0.25.","code":"B"}],"finalAnswers":["|x| < 1/4"],"generatedAt":"2026-05-06T14:04:03.147Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"modulus_inequality","label":"Modulus Inequality Approach","steps":[{"type":"text","order":0,"title":"Recall convergence condition","content":"For a binomial expansion of the form $(1 + u)^n$ where $n$ is not a positive integer, the infinite series is convergent if and only if the absolute value of the variable term is strictly less than $1$:\n\n$$|u| < 1$$"},{"type":"text","order":1,"title":"Set up the inequality","content":"In our expansion, the variable term is $4x$. \n\nSubstituting $u = 4x$ into the convergence condition, we obtain the modulus inequality:\n\n$$|4x| < 1$$"},{"type":"text","order":2,"title":"Solve for x","content":"We can solve the inequality for $|x|$ by dividing both sides by $4$, since $4$ is positive:\n\n$$|x| < \\frac{1}{4}$$\n\nAlternatively, this range of validity can be written as:\n\n$$-\\frac{1}{4} < x < \\frac{1}{4}$$"}]},{"id":"double_inequality","label":"Double Inequality Approach","steps":[{"type":"text","order":0,"title":"Identify the convergence condition","content":"For a binomial expansion of the form $(1 + kx)^n$ where $n$ is not a positive integer, the expansion is valid when the absolute value of the variable term is less than $1$. In this case, the variable term is $4x$, so the condition for validity is:\n\n$$-1 < 4x < 1$$"},{"type":"text","order":1,"title":"Solve the double inequality","content":"To find the range of values for $x$, we divide all parts of the double inequality by $4$:\n\n$$-\\frac{1}{4} < x < \\frac{1}{4}$$"},{"type":"text","order":2,"title":"State the final range","content":"Thus, the binomial expansion is valid for the range of values:\n\n$$-\\frac{1}{4} < x < \\frac{1}{4}$$\n\nThis can also be written in modulus notation as $|x| < \\frac{1}{4}$ or as the open interval $\\left(-\\frac{1}{4}, \\frac{1}{4}\\right)$."}]}],"generatedAt":"2026-06-09T08:10:31.858Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"b49b1f0c-83c4-4007-b93d-0d72866e492b","label":"c","content":"Use your expansion to find the expansion of $\\frac{3x-1}{(1+4x)^2}$ up to and including $x^3$","markscheme":"- Multiplies $(3x-1)$ by the expansion from part (a): $(3x-1)(1 - 8x + 48x^2 - 256x^3)$ M1\n- Expands: $3x - 24x^2 + 144x^3 - 1 + 8x - 48x^2 + 256x^3$ M1\n- Simplifies: $-1 + 11x - 72x^2 + 400x^3$ A1","marks":3,"order":2,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Correct application of multiplication by (3x - 1).","code":"M"},{"id":"p2","marks":1,"description":"Expanding terms accurately up to x^3.","code":"M"},{"id":"p3","marks":1,"description":"Correct final simplified expression -1 + 11x - 72x^2 + 400x^3.","code":"A"}],"finalAnswers":["-1+11x-72x^2+400x^3"],"generatedAt":"2026-05-06T14:04:03.147Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"direct_multiplication","label":"Direct Multiplication","steps":[{"type":"text","order":0,"title":"Set up the multiplication","content":"To expand the expression $\\frac{3x-1}{(1+4x)^2}$, we can write it as a product of a linear expression and a negative binomial term:\n\n$$\\frac{3x-1}{(1+4x)^2} = (3x-1)(1+4x)^{-2}$$\n\nUsing the binomial expansion of $(1+4x)^{-2}$ up to the $x^3$ term, we have:\n\n$$(1+4x)^{-2} \\approx 1 - 8x + 48x^2 - 256x^3$$\n\nNow we set up the product:\n\n$$(3x - 1)(1 - 8x + 48x^2 - 256x^3)$$"},{"type":"text","order":1,"title":"Multiply the terms","content":"We distribute each term of $(3x-1)$ across the expansion of $(1+4x)^{-2}$:\n\n$$(3x-1)(1 - 8x + 48x^2 - 256x^3) = 3x(1 - 8x + 48x^2 - 256x^3) - 1(1 - 8x + 48x^2 - 256x^3)$$\n\n$$= (3x - 24x^2 + 144x^3 - 768x^4) - (1 - 8x + 48x^2 - 256x^3)$$\n\nSince we only need terms up to $x^3$, we can ignore the term with $x^4$ and higher powers."},{"type":"text","order":2,"title":"Collect and simplify","content":"Let's list our terms up to $x^3$ and group them by the power of $x$:\n\n* **Constant term ($x^0$):** $-1$\n* **Linear term ($x^1$):** $3x + 8x = 11x$\n* **Quadratic term ($x^2$):** $-24x^2 - 48x^2 = -72x^2$\n* **Cubic term ($x^3$):** $144x^3 + 256x^3 = 400x^3$\n\nPutting these together, we get:\n\n$$-1 + 11x - 72x^2 + 400x^3$$"}]},{"id":"split_terms","label":"Splitting the Fraction","steps":[{"type":"text","order":0,"title":"Recall the binomial expansion","content":"From part (a) of this question, we have already found the binomial expansion of $(1+4x)^{-2}$ up to and including the $x^3$ term:\n\n$$(1+4x)^{-2} = 1 - 8x + 48x^2 - 256x^3 + \\dots$$"},{"type":"text","order":1,"title":"Split the fraction","content":"We split the given fraction into two separate parts so that we can expand each part individually and then combine them:\n\n$$\\frac{3x-1}{(1+4x)^2} = 3x(1+4x)^{-2} - (1+4x)^{-2}$$"},{"type":"text","order":2,"title":"Expand the first part","content":"Let's expand the first part, $3x(1+4x)^{-2}$. We multiply the known binomial expansion of $(1+4x)^{-2}$ by $3x$:\n\n$$3x(1+4x)^{-2} = 3x(1 - 8x + 48x^2 - 256x^3 + \\dots)$$\n\n$$= 3x - 24x^2 + 144x^3 - 768x^4 + \\dots$$\n\nSince we only need terms up to and including $x^3$, we can ignore the $x^4$ and higher-power terms:\n\n$$3x(1+4x)^{-2} \\approx 3x - 24x^2 + 144x^3$$"},{"type":"text","order":3,"title":"Expand the second part","content":"Now let's expand the second part, $-(1+4x)^{-2}$. We multiply the known binomial expansion of $(1+4x)^{-2}$ by $-1$:\n\n$$-(1+4x)^{-2} = -(1 - 8x + 48x^2 - 256x^3 + \\dots)$$\n\nKeeping terms up to $x^3$, this gives:\n\n$$-(1+4x)^{-2} \\approx -1 + 8x - 48x^2 + 256x^3$$"},{"type":"text","order":4,"title":"Combine and simplify terms","content":"Finally, we add the two expansions together and collect like terms to get our final result:\n\n$$\\frac{3x-1}{(1+4x)^2} = 3x(1+4x)^{-2} - (1+4x)^{-2}$$\n\n$$\\approx (3x - 24x^2 + 144x^3) + (-1 + 8x - 48x^2 + 256x^3)$$\n\n$$= -1 + (3 + 8)x + (-24 - 48)x^2 + (144 + 256)x^3$$\n\n$$= -1 + 11x - 72x^2 + 400x^3$$\n\nThis final simplified expression matches the expected result from the markscheme."}]}],"generatedAt":"2026-06-09T08:12:57.450Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"547ecc9b-6a40-45eb-b47f-0ca1f4a971bc:binomial_expansion","label":"(a) Binomial Series Expansion","steps":[{"type":"text","order":0,"title":"State the formula","content":"To expand $(1+4x)^{-2}$, we use the general binomial series expansion for a rational or negative index $n$:\n\n$$(1+u)^n = 1 + nu + \\frac{n(n-1)}{2!}u^2 + \\frac{n(n-1)(n-2)}{3!}u^3 + \\dots$$\n\nThis formula is valid when $|u| < 1$. Here, we will substitute $u = 4x$ and $n = -2$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u = \\htmlData{anchor=u-val}{\\color{@sky}{4x}}, \\quad n = \\htmlData{anchor=n-val}{\\color{@amber}{-2}}$"},{"latex":"$$(1 + \\htmlData{anchor=u-sub}{\\color{@sky}{4x}})^{\\htmlData{anchor=n-sub}{\\color{@amber}{-2}}} = 1 + (\\htmlData{anchor=n-term1}{\\color{@amber}{-2}})(\\htmlData{anchor=u-term1}{\\color{@sky}{4x}}) + \\frac{(\\htmlData{anchor=n-term2a}{\\color{@amber}{-2}})(\\htmlData{anchor=n-term2b}{\\color{@amber}{-3}})}{2}(\\htmlData{anchor=u-term2}{\\color{@sky}{4x}})^2 + \\frac{(\\htmlData{anchor=n-term3a}{\\color{@amber}{-2}})(\\htmlData{anchor=n-term3b}{\\color{@amber}{-3}})(\\htmlData{anchor=n-term3c}{\\color{@amber}{-4}})}{6}(\\htmlData{anchor=u-term3}{\\color{@sky}{4x}})^3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-val"},{"color":"amber","style":"text-color","anchor":"n-val"}],"connections":[{"to":"u-sub","from":"u-val","color":"sky","label":null,"style":"solid"},{"to":"n-sub","from":"n-val","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Substitute the values","description":"We substitute $u = 4x$ and $n = -2$ into the binomial expansion formula. Writing down this structured substitution earns the first method mark."},{"type":"text","order":2,"title":"Simplify each term","content":"Now we simplify each term step-by-step:\n\n* **$x$ term:**\n $$(-2)(4x) = -8x$$\n\n* **$x^2$ term:**\n $$\\frac{(-2)(-3)}{2}(4x)^2 = \\frac{6}{2}(16x^2) = 3(16x^2) = 48x^2$$\n\n* **$x^3$ term:**\n $$\\frac{(-2)(-3)(-4)}{6}(4x)^3 = \\frac{-24}{6}(64x^3) = -4(64x^3) = -256x^3$$"},{"type":"text","order":3,"title":"State final expansion","content":"Combining all the simplified terms, we get the final expansion up to the $x^3$ term:\n\n$$(1+4x)^{-2} = 1 - 8x + 48x^2 - 256x^3 + \\dots$$\n\nEach correct term earns an accuracy mark in the markscheme."}]},{"id":"547ecc9b-6a40-45eb-b47f-0ca1f4a971bc:maclaurin_series","label":"(a) Maclaurin Series Expansion","steps":[{"type":"text","order":0,"title":"Define the Maclaurin Series","content":"To find the expansion of $f(x) = (1+4x)^{-2}$ up to the term in $x^3$ using the Maclaurin series, we use the standard formula:\n\n$$f(x) = f(0) + f'(0)x + \\frac{f''(0)}{2!}x^2 + \\frac{f'''(0)}{3!}x^3 + \\dots$$\n\nFirst, let's establish our function and evaluate it at $x = 0$:\n\n$$f(x) = (1+4x)^{-2}$$\n\n$$f(0) = (1+0)^{-2} = 1$$"},{"type":"text","order":1,"title":"Find the first derivative","content":"Using the chain rule, we differentiate $f(x) = (1+4x)^{-2}$:\n\n$$f'(x) = -2(1+4x)^{-3} \\cdot \\frac{d}{dx}(1+4x)$$\n\n$$f'(x) = -2(1+4x)^{-3} \\cdot 4 = -8(1+4x)^{-3}$$\n\nNow, evaluate $f'(x)$ at $x = 0$:\n\n$$f'(0) = -8(1+0)^{-3} = -8$$"},{"type":"text","order":2,"title":"Find the second derivative","content":"Next, we differentiate $f'(x) = -8(1+4x)^{-3}$ to find the second derivative:\n\n$$f''(x) = -8 \\cdot (-3)(1+4x)^{-4} \\cdot 4$$\n\n$$f''(x) = 24 \\cdot 4(1+4x)^{-4} = 96(1+4x)^{-4}$$\n\nNow, evaluate $f''(x)$ at $x = 0$:\n\n$$f''(0) = 96(1+0)^{-4} = 96$$"},{"type":"text","order":3,"title":"Find the third derivative","content":"We differentiate $f''(x) = 96(1+4x)^{-4}$ one more time to find the third derivative:\n\n$$f'''(x) = 96 \\cdot (-4)(1+4x)^{-5} \\cdot 4$$\n\n$$f'''(x) = -384 \\cdot 4(1+4x)^{-5} = -1536(1+4x)^{-5}$$\n\nNow, evaluate $f'''(x)$ at $x = 0$:\n\n$$f'''(0) = -1536(1+0)^{-5} = -1536$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(0) = \\htmlData{anchor=val0}{\\color{@sky}{1}}, \\quad f'(0) = \\htmlData{anchor=val1}{\\color{@amber}{-8}}, \\quad f''(0) = \\htmlData{anchor=val2}{\\color{@purple}{96}}, \\quad f'''(0) = \\htmlData{anchor=val3}{\\color{@teal}{-1536}}$"},{"latex":"$$f(x) \\approx \\htmlData{anchor=term0}{\\color{@sky}{1}} + \\htmlData{anchor=term1}{\\color{@amber}{(-8)}}x + \\frac{\\htmlData{anchor=term2}{\\color{@purple}{96}}}{2!}x^2 + \\frac{\\htmlData{anchor=term3}{\\color{@teal}{-1536}}}{3!}x^3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"val0"},{"color":"amber","style":"text-color","anchor":"val1"},{"color":"purple","style":"text-color","anchor":"val2"},{"color":"teal","style":"text-color","anchor":"val3"}],"connections":[{"to":"term0","from":"val0","color":"sky","label":null,"style":"solid"},{"to":"term1","from":"val1","color":"amber","label":null,"style":"solid"},{"to":"term2","from":"val2","color":"purple","label":null,"style":"solid"},{"to":"term3","from":"val3","color":"teal","label":null,"style":"solid"}]},"order":4,"title":"Substitute into Maclaurin Formula","description":"We substitute our calculated values for $f(0)$, $f'(0)$, $f''(0)$, and $f'''(0)$ directly into the Maclaurin series expansion formula."},{"type":"text","order":5,"title":"Simplify terms to final answer","content":"Now, simplify the coefficients to obtain the final expanded expression:\n\n* For the $x^2$ term: \n$$\\frac{96}{2!}x^2 = \\frac{96}{2}x^2 = 48x^2$$\n\n* For the $x^3$ term: \n$$\\frac{-1536}{3!}x^3 = \\frac{-1536}{6}x^3 = -256x^3$$\n\nPutting it all together, we get:\n\n$$(1+4x)^{-2} \\approx 1 - 8x + 48x^2 - 256x^3$$"}]},{"id":"547ecc9b-6a40-45eb-b47f-0ca1f4a971bc:long_division","label":"(a) Algebraic Long Division","steps":[{"type":"text","order":0,"title":"Rewrite the expression","content":"To expand $(1+4x)^{-2}$ using algebraic division, we first rewrite the negative power as a fraction:\n\n$$(1+4x)^{-2} = \\frac{1}{(1+4x)^2}$$","highlights":[{"text":"$$(1+4x)^{-2}$","location":"specification"}]},{"type":"text","order":1,"title":"Expand the denominator","content":"Next, we expand the quadratic expression in the denominator:\n\n$$(1+4x)^2 = 1 + 2(4x) + (4x)^2 = 1 + 8x + 16x^2$$\n\nSubstituting this back into our fraction gives:\n\n$$\\frac{1}{1 + 8x + 16x^2}$$"},{"type":"text","order":2,"title":"Perform algebraic division","content":"To perform the division of $1$ by $1 + 8x + 16x^2$, we divide term-by-term:\n\n1. **First term:** $1 \\div 1 = 1$.\n * Subtracting $1(1 + 8x + 16x^2)$ from $1$ leaves: $$-8x - 16x^2$$\n\n2. **Second term:** $-8x \\div 1 = -8x$.\n * Subtracting $-8x(1 + 8x + 16x^2) = -8x - 64x^2 - 128x^3$ from $-8x - 16x^2$ leaves: $$48x^2 + 128x^3$$\n\n3. **Third term:** $48x^2 \\div 1 = 48x^2$.\n * Subtracting $48x^2(1 + 8x + 16x^2) = 48x^2 + 384x^3 + 768x^4$ from $48x^2 + 128x^3$ leaves: $$-256x^3 - 768x^4$$\n\n4. **Fourth term:** $-256x^3 \\div 1 = -256x^3$.\n\nCombining these quotients yields the series up to the $x^3$ term:\n$$1 - 8x + 48x^2 - 256x^3 + \\dots$$"},{"type":"text","order":3,"title":"State the final expansion","content":"From the quotient of our division, the expansion of $(1+4x)^{-2}$ up to the term in $x^3$ is:\n\n$$1 - 8x + 48x^2 - 256x^3$$"}]},{"id":"3f556779-73ca-4be1-a8f0-b05439815f71:modulus_inequality","label":"(b) Modulus Inequality Approach","steps":[{"type":"text","order":0,"title":"Recall convergence condition","content":"For a binomial expansion of the form $(1 + u)^n$ where $n$ is not a positive integer, the infinite series is convergent if and only if the absolute value of the variable term is strictly less than $1$:\n\n$$|u| < 1$$"},{"type":"text","order":1,"title":"Set up the inequality","content":"In our expansion, the variable term is $4x$. \n\nSubstituting $u = 4x$ into the convergence condition, we obtain the modulus inequality:\n\n$$|4x| < 1$$"},{"type":"text","order":2,"title":"Solve for x","content":"We can solve the inequality for $|x|$ by dividing both sides by $4$, since $4$ is positive:\n\n$$|x| < \\frac{1}{4}$$\n\nAlternatively, this range of validity can be written as:\n\n$$-\\frac{1}{4} < x < \\frac{1}{4}$$"}]},{"id":"3f556779-73ca-4be1-a8f0-b05439815f71:double_inequality","label":"(b) Double Inequality Approach","steps":[{"type":"text","order":0,"title":"Identify the convergence condition","content":"For a binomial expansion of the form $(1 + kx)^n$ where $n$ is not a positive integer, the expansion is valid when the absolute value of the variable term is less than $1$. In this case, the variable term is $4x$, so the condition for validity is:\n\n$$-1 < 4x < 1$$"},{"type":"text","order":1,"title":"Solve the double inequality","content":"To find the range of values for $x$, we divide all parts of the double inequality by $4$:\n\n$$-\\frac{1}{4} < x < \\frac{1}{4}$$"},{"type":"text","order":2,"title":"State the final range","content":"Thus, the binomial expansion is valid for the range of values:\n\n$$-\\frac{1}{4} < x < \\frac{1}{4}$$\n\nThis can also be written in modulus notation as $|x| < \\frac{1}{4}$ or as the open interval $\\left(-\\frac{1}{4}, \\frac{1}{4}\\right)$."}]},{"id":"b49b1f0c-83c4-4007-b93d-0d72866e492b:direct_multiplication","label":"(c) Direct Multiplication","steps":[{"type":"text","order":0,"title":"Set up the multiplication","content":"To expand the expression $\\frac{3x-1}{(1+4x)^2}$, we can write it as a product of a linear expression and a negative binomial term:\n\n$$\\frac{3x-1}{(1+4x)^2} = (3x-1)(1+4x)^{-2}$$\n\nUsing the binomial expansion of $(1+4x)^{-2}$ up to the $x^3$ term, we have:\n\n$$(1+4x)^{-2} \\approx 1 - 8x + 48x^2 - 256x^3$$\n\nNow we set up the product:\n\n$$(3x - 1)(1 - 8x + 48x^2 - 256x^3)$$"},{"type":"text","order":1,"title":"Multiply the terms","content":"We distribute each term of $(3x-1)$ across the expansion of $(1+4x)^{-2}$:\n\n$$(3x-1)(1 - 8x + 48x^2 - 256x^3) = 3x(1 - 8x + 48x^2 - 256x^3) - 1(1 - 8x + 48x^2 - 256x^3)$$\n\n$$= (3x - 24x^2 + 144x^3 - 768x^4) - (1 - 8x + 48x^2 - 256x^3)$$\n\nSince we only need terms up to $x^3$, we can ignore the term with $x^4$ and higher powers."},{"type":"text","order":2,"title":"Collect and simplify","content":"Let's list our terms up to $x^3$ and group them by the power of $x$:\n\n* **Constant term ($x^0$):** $-1$\n* **Linear term ($x^1$):** $3x + 8x = 11x$\n* **Quadratic term ($x^2$):** $-24x^2 - 48x^2 = -72x^2$\n* **Cubic term ($x^3$):** $144x^3 + 256x^3 = 400x^3$\n\nPutting these together, we get:\n\n$$-1 + 11x - 72x^2 + 400x^3$$"}]},{"id":"b49b1f0c-83c4-4007-b93d-0d72866e492b:split_terms","label":"(c) Splitting the Fraction","steps":[{"type":"text","order":0,"title":"Recall the binomial expansion","content":"From part (a) of this question, we have already found the binomial expansion of $(1+4x)^{-2}$ up to and including the $x^3$ term:\n\n$$(1+4x)^{-2} = 1 - 8x + 48x^2 - 256x^3 + \\dots$$"},{"type":"text","order":1,"title":"Split the fraction","content":"We split the given fraction into two separate parts so that we can expand each part individually and then combine them:\n\n$$\\frac{3x-1}{(1+4x)^2} = 3x(1+4x)^{-2} - (1+4x)^{-2}$$"},{"type":"text","order":2,"title":"Expand the first part","content":"Let's expand the first part, $3x(1+4x)^{-2}$. We multiply the known binomial expansion of $(1+4x)^{-2}$ by $3x$:\n\n$$3x(1+4x)^{-2} = 3x(1 - 8x + 48x^2 - 256x^3 + \\dots)$$\n\n$$= 3x - 24x^2 + 144x^3 - 768x^4 + \\dots$$\n\nSince we only need terms up to and including $x^3$, we can ignore the $x^4$ and higher-power terms:\n\n$$3x(1+4x)^{-2} \\approx 3x - 24x^2 + 144x^3$$"},{"type":"text","order":3,"title":"Expand the second part","content":"Now let's expand the second part, $-(1+4x)^{-2}$. We multiply the known binomial expansion of $(1+4x)^{-2}$ by $-1$:\n\n$$-(1+4x)^{-2} = -(1 - 8x + 48x^2 - 256x^3 + \\dots)$$\n\nKeeping terms up to $x^3$, this gives:\n\n$$-(1+4x)^{-2} \\approx -1 + 8x - 48x^2 + 256x^3$$"},{"type":"text","order":4,"title":"Combine and simplify terms","content":"Finally, we add the two expansions together and collect like terms to get our final result:\n\n$$\\frac{3x-1}{(1+4x)^2} = 3x(1+4x)^{-2} - (1+4x)^{-2}$$\n\n$$\\approx (3x - 24x^2 + 144x^3) + (-1 + 8x - 48x^2 + 256x^3)$$\n\n$$= -1 + (3 + 8)x + (-24 - 48)x^2 + (144 + 256)x^3$$\n\n$$= -1 + 11x - 72x^2 + 400x^3$$\n\nThis final simplified expression matches the expected result from the 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3.4 Sequences and Series (A-level only)

Question 66

3.4 Sequences and Series (A-level only) Questions

3.4 Sequences and Series (A-level only)

Question 66

3.4 Sequences and Series (A-level only) Questions