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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":8,"initialQuestionWindow":[{"id":"3950bc86-f600-4c92-a793-037add492875","revisionId":"206756be-b37d-48d1-90c2-596cc3f46e08","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"b15e6ffe-6846-4366-81eb-734896d2d019","label":"a","content":"Expand $(2+x)^{\\frac{1}{2}}$ in ascending powers of $x$ up to and including $x^3$.","markscheme":"- Factorises: $2^{\\frac{1}{2}}(1 + \\frac{1}{2}x)^{\\frac{1}{2}}$ M1\n- Expands: $\\sqrt{2} [ 1 + (\\frac{1}{2})(\\frac{1}{2}x) + \\frac{(\\frac{1}{2})(-\\frac{1}{2})}{2}(\\frac{1}{2}x)^2 + \\frac{(\\frac{1}{2})(-\\frac{1}{2})(-\\frac{3}{2})}{6}(\\frac{1}{2}x)^3 ]$ M1\n- Simplifies bracket: $1 + \\frac{1}{4}x - \\frac{1}{32}x^2 + \\frac{1}{128}x^3$ A1\n- Final answer: $\\sqrt{2} + \\frac{\\sqrt{2}}{4}x - \\frac{\\sqrt{2}}{32}x^2 + \\frac{\\sqrt{2}}{128}x^3$ A1","marks":4,"order":0,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Factorises out $2^{\\frac{1}{2}}$ correctly.","code":"M"},{"id":"p2","marks":1,"description":"Applies the binomial expansion formula for a fractional power with the correct substituted term $(\\frac{1}{2}x)$.","code":"M"},{"id":"p3","marks":1,"description":"Correctly simplifies the terms inside the bracket up to $x^3$.","code":"A"},{"id":"p4","marks":1,"description":"Correct final expansion with $\\sqrt{2}$ factor distributed.","code":"A"}],"finalAnswers":["\\sqrt{2} + \\frac{\\sqrt{2}}{4}x - \\frac{\\sqrt{2}}{32}x^2 + \\frac{\\sqrt{2}}{128}x^3"],"generatedAt":"2026-05-06T14:04:26.418Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"binomial_factoring","label":"Binomial Theorem with Factoring","steps":[{"type":"text","order":0,"title":"Factorise the expression","content":"To expand $(2+x)^{\\frac{1}{2}}$ using the standard binomial expansion, we must first rewrite it in the form $(1+y)^n$. We do this by factoring out the constant $2$ from inside the brackets:\n\n$$(2+x)^{\\frac{1}{2}} = \\left[2\\left(1 + \\frac{x}{2}\\right)\\right]^{\\frac{1}{2}}$$\n\n$$\\phantom{(2+x)^{\\frac{1}{2}}} = 2^{\\frac{1}{2}}\\left(1 + \\frac{x}{2}\\right)^{\\frac{1}{2}}$$\n\n$$\\phantom{(2+x)^{\\frac{1}{2}}} = \\sqrt{2}\\left(1 + \\frac{x}{2}\\right)^{\\frac{1}{2}}$$\n\nThis step is crucial because the standard binomial expansion formula $(1+y)^n$ is only valid when the first term is $1$. Factoring out $2^{\\frac{1}{2}}$ earns you the first method mark (**M1**)."},{"type":"text","order":1,"title":"Recall the binomial formula","content":"The standard binomial expansion formula for any rational power $n$ is:\n\n$$(1+y)^n = 1 + ny + \\frac{n(n-1)}{2!}y^2 + \\frac{n(n-1)(n-2)}{3!}y^3 + \\dots$$\n\nIn our case, we have:\n- $n = \\frac{1}{2}$\n- $y = \\frac{x}{2}$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\left(1 + \\htmlData{anchor=y-val}{\\color{@sky}{\\frac{x}{2}}}\\right)^{\\htmlData{anchor=n-val}{\\color{@amber}{\\frac{1}{2}}}}$"},{"latex":"$$= 1 + \\left(\\htmlData{anchor=n-sub1}{\\color{@amber}{\\frac{1}{2}}}\\right)\\left(\\htmlData{anchor=y-sub1}{\\color{@sky}{\\frac{x}{2}}}\\right) + \\frac{\\left(\\htmlData{anchor=n-sub2}{\\color{@amber}{\\frac{1}{2}}}\\right)\\left(\\color{@amber}{-\\frac{1}{2}}\\right)}{2}\\left(\\htmlData{anchor=y-sub2}{\\color{@sky}{\\frac{x}{2}}}\\right)^2 + \\frac{\\left(\\htmlData{anchor=n-sub3}{\\color{@amber}{\\frac{1}{2}}}\\right)\\left(\\color{@amber}{-\\frac{1}{2}}\\right)\\left(\\color{@amber}{-\\frac{3}{2}}\\right)}{6}\\left(\\htmlData{anchor=y-sub3}{\\color{@sky}{\\frac{x}{2}}}\\right)^3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-val"},{"color":"amber","style":"text-color","anchor":"n-val"}],"connections":[{"to":"y-sub1","from":"y-val","color":"sky","label":"substitute y","style":"solid"},{"to":"n-sub1","from":"n-val","color":"amber","label":"substitute n","style":"solid"}]},"order":2,"title":"Apply the expansion formula","description":"Substitute $n = \\frac{1}{2}$ and $y = \\frac{x}{2}$ into the general formula up to the $x^3$ term. This setup earns the second method mark (**M1**)."},{"type":"text","order":3,"title":"Simplify inside the bracket","content":"Now, we simplify each term inside the bracket individually:\n\n- **First-degree term:**\n $$\\left(\\frac{1}{2}\\right)\\left(\\frac{x}{2}\\right) = \\frac{1}{4}x$$\n\n- **Second-degree term:**\n $$\\frac{\\left(\\frac{1}{2}\\right)\\left(-\\frac{1}{2}\\right)}{2}\\left(\\frac{x^2}{4}\\right) = \\frac{-\\frac{1}{4}}{2}\\left(\\frac{x^2}{4}\\right) = -\\frac{1}{8}\\left(\\frac{x^2}{4}\\right) = -\\frac{1}{32}x^2$$\n\n- **Third-degree term:**\n $$\\frac{\\left(\\frac{1}{2}\\right)\\left(-\\frac{1}{2}\\right)\\left(-\\frac{3}{2}\\right)}{6}\\left(\\frac{x^3}{8}\\right) = \\frac{\\frac{3}{8}}{6}\\left(\\frac{x^3}{8}\\right) = \\frac{1}{16}\\left(\\frac{x^3}{8}\\right) = \\frac{1}{128}x^3$$\n\nPutting these simplified terms back into the bracket gives:\n\n$$\\left(1 + \\frac{x}{2}\\right)^{\\frac{1}{2}} = 1 + \\frac{1}{4}x - \\frac{1}{32}x^2 + \\frac{1}{128}x^3$$\n\nThis simplified bracket earns an accuracy mark (**A1**)."},{"type":"text","order":4,"title":"Multiply by the factor","content":"Finally, we multiply our simplified expansion by the factor of $\\sqrt{2}$ that we pulled out in Step 1:\n\n$$(2+x)^{\\frac{1}{2}} = \\sqrt{2}\\left(1 + \\frac{1}{4}x - \\frac{1}{32}x^2 + \\frac{1}{128}x^3\\right)$$\n\nSubstituting this multiplication across the expression gives:\n\n$$\\sqrt{2} + \\frac{\\sqrt{2}}{4}x - \\frac{\\sqrt{2}}{32}x^2 + \\frac{\\sqrt{2}}{128}x^3$$\n\nThis completed expression completes Part 1 of the question and earns the final accuracy mark (**A1**)."},{"type":"text","order":5,"title":"Determine the validity range","content":"For Part 2, the binomial expansion of $(1+y)^n$ is valid when $|y| < 1$.\n\nIn our factored expression, we have $y = \\frac{x}{2}$. Therefore, the validity condition is:\n\n$$\\left|\\frac{x}{2}\\right| < 1$$\n\nMultiplying both sides by $2$ gives:\n\n$$|x| < 2$$\n\nThis completes Part 2 and earns (**B1**)."},{"type":"text","order":6,"title":"Relate substitution to root","content":"For Part 3, substituting $x=0.5$ into $(2+x)^{\\frac{1}{2}}$ gives:\n\n$$(2+0.5)^{\\frac{1}{2}} = (2.5)^{\\frac{1}{2}} = \\sqrt{\\frac{5}{2}} = \\frac{\\sqrt{5}}{\\sqrt{2}}$$\n\nTo write this in terms of $\\sqrt{10}$, we multiply the numerator and denominator by $\\sqrt{2}$:\n\n$$\\frac{\\sqrt{5} \\times \\sqrt{2}}{\\sqrt{2} \\times \\sqrt{2}} = \\frac{\\sqrt{10}}{2}$$\n\nThis means that when we evaluate our binomial expansion at $x = 0.5$, we will be approximating $\\frac{\\sqrt{10}}{2}$ (**M1**)."},{"type":"text","order":7,"title":"Approximate the square root","content":"Now, we substitute $x=0.5$ into our series expansion from Part 1:\n\n$$\\frac{\\sqrt{10}}{2} \\approx \\sqrt{2} + \\frac{\\sqrt{2}}{4}(0.5) - \\frac{\\sqrt{2}}{32}(0.5)^2 + \\frac{\\sqrt{2}}{128}(0.5)^3$$\n\nLet's evaluate this numerical expression:\n\n$$\\frac{\\sqrt{10}}{2} \\approx \\sqrt{2} + \\frac{\\sqrt{2}}{8} - \\frac{\\sqrt{2}}{128} + \\frac{\\sqrt{2}}{1024}$$\n\n$$\\frac{\\sqrt{10}}{2} \\approx \\sqrt{2}\\left(1 + \\frac{1}{8} - \\frac{1}{128} + \\frac{1}{1024}\\right)$$\n\n$$\\frac{\\sqrt{10}}{2} \\approx \\sqrt{2}\\left(\\frac{1145}{1024}\\right) \\approx 1.58132...\\text{ (**M1**)}$$\n\nMultiplying by $2$ to solve for $\\sqrt{10}$:\n\n$$\\sqrt{10} \\approx 2 \\times 1.58132... = 3.16264...$$\n\nRounding to 3 decimal places gives:\n\n$$\\sqrt{10} \\approx 3.163\\text{ (**A1**)}$$"}]},{"id":"maclaurin_series","label":"Maclaurin Series Expansion","steps":[{"type":"text","order":0,"title":"State the Maclaurin formula","content":"To expand $f(x) = (2+x)^{\\frac{1}{2}}$ using the Maclaurin series up to the $x^3$ term, we use the formula:\n\n$$f(x) \\approx f(0) + f'(0)x + \\frac{f''(0)}{2!}x^2 + \\frac{f'''(0)}{3!}x^3$$\n\nFirst, we need to find the function value and its first three derivatives evaluated at $x = 0$."},{"type":"text","order":1,"title":"Find the function value","content":"For our function, we have:\n\n$$f(x) = (2+x)^{\\frac{1}{2}}$$\n\nEvaluating this at $x = 0$ gives:\n\n$$f(0) = (2+0)^{\\frac{1}{2}} = \\sqrt{2}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f'(x) = \\htmlData{anchor=pow-1}{\\frac{1}{2}}(2+x)^{\\htmlData{anchor=new-pow-1}{-\\frac{1}{2}}}$"},{"latex":"$$f'(0) = \\frac{1}{2}(2+0)^{-\\frac{1}{2}} = \\frac{1}{2\\sqrt{2}} = \\htmlData{anchor=f-prime-final}{\\frac{\\sqrt{2}}{4}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"pow-1"},{"color":"green","style":"box","anchor":"f-prime-final"}],"connections":[{"to":"new-pow-1","from":"pow-1","color":"sky","label":"Power rule","style":"solid"}]},"order":2,"title":"Evaluate the first derivative","description":"We apply the power rule to differentiate $f(x)$ with respect to $x$ and then evaluate it at $x=0$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f''(x) = \\left(\\frac{1}{2}\\right)\\left(\\htmlData{anchor=pow-2}{-\\frac{1}{2}}\\right)(2+x)^{\\htmlData{anchor=new-pow-2}{-\\frac{3}{2}}} = -\\frac{1}{4}(2+x)^{-\\frac{3}{2}}$"},{"latex":"$$f''(0) = -\\frac{1}{4}(2)^{-\\frac{3}{2}} = -\\frac{1}{4}\\left(\\frac{1}{2\\sqrt{2}}\\right) = \\htmlData{anchor=f-double-prime-final}{-\\frac{\\sqrt{2}}{16}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"pow-2"},{"color":"green","style":"box","anchor":"f-double-prime-final"}],"connections":[{"to":"new-pow-2","from":"pow-2","color":"amber","label":"Power rule","style":"solid"}]},"order":3,"title":"Evaluate the second derivative","description":"Differentiating once more gives the second derivative."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f'''(x) = \\left(-\\frac{1}{4}\\right)\\left(\\htmlData{anchor=pow-3}{-\\frac{3}{2}}\\right)(2+x)^{\\htmlData{anchor=new-pow-3}{-\\frac{5}{2}}} = \\frac{3}{8}(2+x)^{-\\frac{5}{2}}$"},{"latex":"$$f'''(0) = \\frac{3}{8}(2)^{-\\frac{5}{2}} = \\frac{3}{8}\\left(\\frac{1}{4\\sqrt{2}}\\right) = \\htmlData{anchor=f-triple-prime-final}{\\frac{3\\sqrt{2}}{64}}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"pow-3"},{"color":"green","style":"box","anchor":"f-triple-prime-final"}],"connections":[{"to":"new-pow-3","from":"pow-3","color":"purple","label":"Power rule","style":"solid"}]},"order":4,"title":"Evaluate the third derivative","description":"Finally, we calculate the third derivative and its value at $x = 0$."},{"type":"text","order":5,"title":"Construct the Maclaurin series","content":"Now we substitute our values back into the Maclaurin series formula:\n\n$$f(x) \\approx f(0) + f'(0)x + \\frac{f''(0)}{2!}x^2 + \\frac{f'''(0)}{3!}x^3$$\n\nSubstituting the values we calculated:\n\n$$f(x) \\approx \\sqrt{2} + \\left(\\frac{\\sqrt{2}}{4}\\right)x + \\frac{-\\frac{\\sqrt{2}}{16}}{2}x^2 + \\frac{\\frac{3\\sqrt{2}}{64}}{6}x^3$$\n\nSimplifying each coefficient:\n\n- **For the $x^2$ term:**\n $$\\frac{-\\frac{\\sqrt{2}}{16}}{2} = -\\frac{\\sqrt{2}}{32}$$\n\n- **For the $x^3$ term:**\n $$\\frac{\\frac{3\\sqrt{2}}{64}}{6} = \\frac{\\sqrt{2}}{128}$$\n\nThis gives our final series expansion:\n\n$$\\sqrt{2} + \\frac{\\sqrt{2}}{4}x - \\frac{\\sqrt{2}}{32}x^2 + \\frac{\\sqrt{2}}{128}x^3$$"},{"type":"text","order":6,"title":"Determine the validity range","content":"For Part 2, the range of convergence of the series is based on the distance to the nearest singularity in the complex plane (at $x = -2$), or equivalently, using the binomial series condition $|y| < 1$ where $y = \\frac{x}{2}$:\n\n$$\\left|\\frac{x}{2}\\right| < 1 \\implies |x| < 2$$\n\nThis completes Part 2 and earns (**B1**)."},{"type":"text","order":7,"title":"Relate substitution to root","content":"For Part 3, we substitute $x = 0.5$ into our function $f(x) = (2+x)^{\\frac{1}{2}}$:\n\n$$f(0.5) = (2 + 0.5)^{\\frac{1}{2}} = \\sqrt{2.5} = \\sqrt{\\frac{5}{2}} = \\frac{\\sqrt{5}}{\\sqrt{2}} = \\frac{\\sqrt{10}}{2}$$\n\nThus, evaluating the series at $x = 0.5$ provides an approximation for $\\frac{\\sqrt{10}}{2}$ (**M1**)."},{"type":"text","order":8,"title":"Approximate the square root","content":"Substituting $x = 0.5$ into the series expansion we obtained:\n\n$$\\frac{\\sqrt{10}}{2} \\approx \\sqrt{2} + \\frac{\\sqrt{2}}{4}(0.5) - \\frac{\\sqrt{2}}{32}(0.5)^2 + \\frac{\\sqrt{2}}{128}(0.5)^3 \\approx 1.58132...\\text{ (**M1**)}$$\n\nMultiplying by $2$ to isolate $\\sqrt{10}$:\n\n$$\\sqrt{10} \\approx 2 \\times 1.58132... = 3.16264...$$\n\nRounding to 3 decimal places:\n\n$$\\sqrt{10} \\approx 3.163\\text{ (**A1**)}$$"}]}],"generatedAt":"2026-06-09T04:36:07.876Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"75cc722a-d82e-4195-9a67-d1d5cd5b4f57","label":"b","content":"Write down the range of values of $x$ for which the expansion is valid.","markscheme":"- $|\\frac{1}{2}x| < 1 \\Rightarrow |x| < 2$ B1","marks":1,"order":1,"markingRubric":{"totalMarks":1,"points":[{"id":"p1","marks":1,"description":"States the valid range for the expansion.","code":"B"}],"finalAnswers":["|x| < 2"],"generatedAt":"2026-05-06T14:04:26.418Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"absolute_value_inequality","label":"Absolute Value Inequality","steps":[{"type":"text","order":0,"title":"Recall the validity condition","content":"For a binomial expansion of the form $(1 + u)^n$ where $n$ is a non-integer, the expansion is valid (converges) if and only if the absolute value of the variable term, $u$, is strictly less than $1$: $$|u| < 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u = \\htmlData{anchor=u-term}{\\color{@sky}{\\frac{1}{2}x}}$"},{"latex":"$$\\left|\\htmlData{anchor=u-ineq}{\\color{@sky}{u}}\\right| < 1$"},{"latex":"$$\\left|\\htmlData{anchor=sub-ineq}{\\color{@sky}{\\frac{1}{2}x}}\\right| < 1$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"sky","style":"text-color","anchor":"sub-ineq"}],"connections":[{"to":"sub-ineq","from":"u-term","color":"sky","label":"substitute","style":"solid"},{"to":"sub-ineq","from":"u-ineq","color":"sky","label":null,"style":"dashed"}]},"order":1,"title":"Substitute the variable term","description":"In this expansion, our variable term is $u = \\frac{1}{2}x$. We set up the absolute value inequality using this term."},{"type":"text","order":2,"title":"Solve the inequality","content":"To find the range of values for $x$, we solve the inequality: $$\\left|\\frac{1}{2}x\\right| < 1$$ $$\\frac{1}{2}|x| < 1$$ $$|x| < 2$$ This can also be written as $-2 < x < 2$. Both formats represent the range of values for which the expansion is valid."}]},{"id":"interval_inequality","label":"Interval Inequality","steps":[{"type":"text","order":0,"title":"Recall the validity condition","content":"For a binomial expansion of $(1 + y)^n$ where $n$ is a fraction or negative, the expansion is only valid when the variable term $y$ lies strictly between $-1$ and $1$. In our case, the variable term in the bracket is $\\frac{1}{2}x$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Identify the term: } \\htmlData{anchor=term}{\\frac{1}{2}x}$"},{"latex":"$$-1 < \\htmlData{anchor=ineq}{\\frac{1}{2}x} < 1$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"term"},{"color":"sky","style":"text-color","anchor":"ineq"}],"connections":[{"to":"ineq","from":"term","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Set up the inequality","description":"We substitute our term, $\\frac{1}{2}x$, into the standard validity double inequality $-1 < y < 1$."},{"type":"text","order":2,"title":"Solve the double inequality","content":"To isolate $x$, we multiply all parts of the double inequality by $2$: $$-1 \\times 2 < x < 1 \\times 2$$ $$-2 < x < 2$$ This gives us the open interval $-2 < x < 2$. This can also be expressed in modulus notation as $|x| < 2$. Either form is fully correct and earns the B1 mark."}]}],"generatedAt":"2026-06-09T04:37:34.420Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"4417af3b-ab13-4538-87dc-3673682081c8","label":"c","content":"Use your expansion, with $x=0.5$, to find an approximation of the value of $\\sqrt{10}$. Give your answer to 3 decimal places.","markscheme":"- Substitutes $x=0.5$ into $(2+x)^{\\frac{1}{2}}$ to get $(2.5)^{\\frac{1}{2}} = \\sqrt{\\frac{5}{2}} = \\frac{\\sqrt{5}}{\\sqrt{2}} = \\frac{\\sqrt{10}}{2}$ M1\n- Substitutes $x=0.5$ into expansion: $\\frac{\\sqrt{10}}{2} \\approx 1.58132...$ M1\n- Solves for $\\sqrt{10} = 2 \\times 1.58132... = 3.163$ A1","marks":3,"order":2,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Relates the substituted value $(2.5)^{\\frac{1}{2}}$ to $\\sqrt{10}$ correctly.","code":"M"},{"id":"p2","marks":1,"description":"Correctly substitutes $x=0.5$ into the expansion found in part (a).","code":"M"},{"id":"p3","marks":1,"description":"Obtains the final approximation of $\\sqrt{10}$ rounded to 3 decimal places.","code":"A"}],"finalAnswers":["3.163"],"generatedAt":"2026-05-06T14:04:26.418Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"direct_substitution","label":"Direct Substitution and Algebraic Manipulation","steps":[{"type":"text","order":0,"title":"Relate $(2.5)^{\\frac{1}{2}}$ to $\\sqrt{10}$","content":"First, we substitute $x = 0.5$ into the original expression, $(2 + x)^{\\frac{1}{2}}$. This gives:\n\n$$(2 + 0.5)^{\\frac{1}{2}} = (2.5)^{\\frac{1}{2}} = \\sqrt{\\frac{5}{2}}$$\n\nTo relate this to $\\sqrt{10}$, we can rationalize the denominator by multiplying the numerator and denominator inside the square root by $2$:\n\n$$\\sqrt{\\frac{5}{2}} = \\sqrt{\\frac{5 \\times 2}{2 \\times 2}} = \\sqrt{\\frac{10}{4}} = \\frac{\\sqrt{10}}{2}$$\n\nThis shows us that:\n\n$$\\frac{\\sqrt{10}}{2} = (2 + 0.5)^{\\frac{1}{2}}$$"},{"type":"text","order":1,"title":"Evaluate the expansion","content":"Next, we substitute $x = 0.5$ into the binomial expansion of $(2 + x)^{\\frac{1}{2}}$ that was found in the previous part of the question. \n\nEvaluating the series expansion up to the given terms at $x = 0.5$ yields:\n\n$$(2 + 0.5)^{\\frac{1}{2}} \\approx 1.58132...$$\n\nSince we established that $(2 + 0.5)^{\\frac{1}{2}} = \\frac{\\sqrt{10}}{2}$, we can write:\n\n$$\\frac{\\sqrt{10}}{2} \\approx 1.58132...$$"},{"type":"text","order":2,"title":"Solve for $\\sqrt{10}$","content":"To find the approximation for $\\sqrt{10}$, we multiply both sides of our relationship by $2$:\n\n$$\\sqrt{10} \\approx 2 \\times 1.58132...$$\n\n$$\\sqrt{10} \\approx 3.16264...$$\n\nRounding this to $3$ decimal places, we get:\n\n$$\\sqrt{10} \\approx 3.163$$"}]}],"generatedAt":"2026-06-09T04:39:05.423Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"b15e6ffe-6846-4366-81eb-734896d2d019:binomial_factoring","label":"(a) Binomial Theorem with Factoring","steps":[{"type":"text","order":0,"title":"Factorise the expression","content":"To expand $(2+x)^{\\frac{1}{2}}$ using the standard binomial expansion, we must first rewrite it in the form $(1+y)^n$. We do this by factoring out the constant $2$ from inside the brackets:\n\n$$(2+x)^{\\frac{1}{2}} = \\left[2\\left(1 + \\frac{x}{2}\\right)\\right]^{\\frac{1}{2}}$$\n\n$$\\phantom{(2+x)^{\\frac{1}{2}}} = 2^{\\frac{1}{2}}\\left(1 + \\frac{x}{2}\\right)^{\\frac{1}{2}}$$\n\n$$\\phantom{(2+x)^{\\frac{1}{2}}} = \\sqrt{2}\\left(1 + \\frac{x}{2}\\right)^{\\frac{1}{2}}$$\n\nThis step is crucial because the standard binomial expansion formula $(1+y)^n$ is only valid when the first term is $1$. Factoring out $2^{\\frac{1}{2}}$ earns you the first method mark (**M1**)."},{"type":"text","order":1,"title":"Recall the binomial formula","content":"The standard binomial expansion formula for any rational power $n$ is:\n\n$$(1+y)^n = 1 + ny + \\frac{n(n-1)}{2!}y^2 + \\frac{n(n-1)(n-2)}{3!}y^3 + \\dots$$\n\nIn our case, we have:\n- $n = \\frac{1}{2}$\n- $y = \\frac{x}{2}$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\left(1 + \\htmlData{anchor=y-val}{\\color{@sky}{\\frac{x}{2}}}\\right)^{\\htmlData{anchor=n-val}{\\color{@amber}{\\frac{1}{2}}}}$"},{"latex":"$$= 1 + \\left(\\htmlData{anchor=n-sub1}{\\color{@amber}{\\frac{1}{2}}}\\right)\\left(\\htmlData{anchor=y-sub1}{\\color{@sky}{\\frac{x}{2}}}\\right) + \\frac{\\left(\\htmlData{anchor=n-sub2}{\\color{@amber}{\\frac{1}{2}}}\\right)\\left(\\color{@amber}{-\\frac{1}{2}}\\right)}{2}\\left(\\htmlData{anchor=y-sub2}{\\color{@sky}{\\frac{x}{2}}}\\right)^2 + \\frac{\\left(\\htmlData{anchor=n-sub3}{\\color{@amber}{\\frac{1}{2}}}\\right)\\left(\\color{@amber}{-\\frac{1}{2}}\\right)\\left(\\color{@amber}{-\\frac{3}{2}}\\right)}{6}\\left(\\htmlData{anchor=y-sub3}{\\color{@sky}{\\frac{x}{2}}}\\right)^3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-val"},{"color":"amber","style":"text-color","anchor":"n-val"}],"connections":[{"to":"y-sub1","from":"y-val","color":"sky","label":"substitute y","style":"solid"},{"to":"n-sub1","from":"n-val","color":"amber","label":"substitute n","style":"solid"}]},"order":2,"title":"Apply the expansion formula","description":"Substitute $n = \\frac{1}{2}$ and $y = \\frac{x}{2}$ into the general formula up to the $x^3$ term. This setup earns the second method mark (**M1**)."},{"type":"text","order":3,"title":"Simplify inside the bracket","content":"Now, we simplify each term inside the bracket individually:\n\n- **First-degree term:**\n $$\\left(\\frac{1}{2}\\right)\\left(\\frac{x}{2}\\right) = \\frac{1}{4}x$$\n\n- **Second-degree term:**\n $$\\frac{\\left(\\frac{1}{2}\\right)\\left(-\\frac{1}{2}\\right)}{2}\\left(\\frac{x^2}{4}\\right) = \\frac{-\\frac{1}{4}}{2}\\left(\\frac{x^2}{4}\\right) = -\\frac{1}{8}\\left(\\frac{x^2}{4}\\right) = -\\frac{1}{32}x^2$$\n\n- **Third-degree term:**\n $$\\frac{\\left(\\frac{1}{2}\\right)\\left(-\\frac{1}{2}\\right)\\left(-\\frac{3}{2}\\right)}{6}\\left(\\frac{x^3}{8}\\right) = \\frac{\\frac{3}{8}}{6}\\left(\\frac{x^3}{8}\\right) = \\frac{1}{16}\\left(\\frac{x^3}{8}\\right) = \\frac{1}{128}x^3$$\n\nPutting these simplified terms back into the bracket gives:\n\n$$\\left(1 + \\frac{x}{2}\\right)^{\\frac{1}{2}} = 1 + \\frac{1}{4}x - \\frac{1}{32}x^2 + \\frac{1}{128}x^3$$\n\nThis simplified bracket earns an accuracy mark (**A1**)."},{"type":"text","order":4,"title":"Multiply by the factor","content":"Finally, we multiply our simplified expansion by the factor of $\\sqrt{2}$ that we pulled out in Step 1:\n\n$$(2+x)^{\\frac{1}{2}} = \\sqrt{2}\\left(1 + \\frac{1}{4}x - \\frac{1}{32}x^2 + \\frac{1}{128}x^3\\right)$$\n\nSubstituting this multiplication across the expression gives:\n\n$$\\sqrt{2} + \\frac{\\sqrt{2}}{4}x - \\frac{\\sqrt{2}}{32}x^2 + \\frac{\\sqrt{2}}{128}x^3$$\n\nThis completed expression completes Part 1 of the question and earns the final accuracy mark (**A1**)."},{"type":"text","order":5,"title":"Determine the validity range","content":"For Part 2, the binomial expansion of $(1+y)^n$ is valid when $|y| < 1$.\n\nIn our factored expression, we have $y = \\frac{x}{2}$. Therefore, the validity condition is:\n\n$$\\left|\\frac{x}{2}\\right| < 1$$\n\nMultiplying both sides by $2$ gives:\n\n$$|x| < 2$$\n\nThis completes Part 2 and earns (**B1**)."},{"type":"text","order":6,"title":"Relate substitution to root","content":"For Part 3, substituting $x=0.5$ into $(2+x)^{\\frac{1}{2}}$ gives:\n\n$$(2+0.5)^{\\frac{1}{2}} = (2.5)^{\\frac{1}{2}} = \\sqrt{\\frac{5}{2}} = \\frac{\\sqrt{5}}{\\sqrt{2}}$$\n\nTo write this in terms of $\\sqrt{10}$, we multiply the numerator and denominator by $\\sqrt{2}$:\n\n$$\\frac{\\sqrt{5} \\times \\sqrt{2}}{\\sqrt{2} \\times \\sqrt{2}} = \\frac{\\sqrt{10}}{2}$$\n\nThis means that when we evaluate our binomial expansion at $x = 0.5$, we will be approximating $\\frac{\\sqrt{10}}{2}$ (**M1**)."},{"type":"text","order":7,"title":"Approximate the square root","content":"Now, we substitute $x=0.5$ into our series expansion from Part 1:\n\n$$\\frac{\\sqrt{10}}{2} \\approx \\sqrt{2} + \\frac{\\sqrt{2}}{4}(0.5) - \\frac{\\sqrt{2}}{32}(0.5)^2 + \\frac{\\sqrt{2}}{128}(0.5)^3$$\n\nLet's evaluate this numerical expression:\n\n$$\\frac{\\sqrt{10}}{2} \\approx \\sqrt{2} + \\frac{\\sqrt{2}}{8} - \\frac{\\sqrt{2}}{128} + \\frac{\\sqrt{2}}{1024}$$\n\n$$\\frac{\\sqrt{10}}{2} \\approx \\sqrt{2}\\left(1 + \\frac{1}{8} - \\frac{1}{128} + \\frac{1}{1024}\\right)$$\n\n$$\\frac{\\sqrt{10}}{2} \\approx \\sqrt{2}\\left(\\frac{1145}{1024}\\right) \\approx 1.58132...\\text{ (**M1**)}$$\n\nMultiplying by $2$ to solve for $\\sqrt{10}$:\n\n$$\\sqrt{10} \\approx 2 \\times 1.58132... = 3.16264...$$\n\nRounding to 3 decimal places gives:\n\n$$\\sqrt{10} \\approx 3.163\\text{ (**A1**)}$$"}]},{"id":"b15e6ffe-6846-4366-81eb-734896d2d019:maclaurin_series","label":"(a) Maclaurin Series Expansion","steps":[{"type":"text","order":0,"title":"State the Maclaurin formula","content":"To expand $f(x) = (2+x)^{\\frac{1}{2}}$ using the Maclaurin series up to the $x^3$ term, we use the formula:\n\n$$f(x) \\approx f(0) + f'(0)x + \\frac{f''(0)}{2!}x^2 + \\frac{f'''(0)}{3!}x^3$$\n\nFirst, we need to find the function value and its first three derivatives evaluated at $x = 0$."},{"type":"text","order":1,"title":"Find the function value","content":"For our function, we have:\n\n$$f(x) = (2+x)^{\\frac{1}{2}}$$\n\nEvaluating this at $x = 0$ gives:\n\n$$f(0) = (2+0)^{\\frac{1}{2}} = \\sqrt{2}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f'(x) = \\htmlData{anchor=pow-1}{\\frac{1}{2}}(2+x)^{\\htmlData{anchor=new-pow-1}{-\\frac{1}{2}}}$"},{"latex":"$$f'(0) = \\frac{1}{2}(2+0)^{-\\frac{1}{2}} = \\frac{1}{2\\sqrt{2}} = \\htmlData{anchor=f-prime-final}{\\frac{\\sqrt{2}}{4}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"pow-1"},{"color":"green","style":"box","anchor":"f-prime-final"}],"connections":[{"to":"new-pow-1","from":"pow-1","color":"sky","label":"Power rule","style":"solid"}]},"order":2,"title":"Evaluate the first derivative","description":"We apply the power rule to differentiate $f(x)$ with respect to $x$ and then evaluate it at $x=0$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f''(x) = \\left(\\frac{1}{2}\\right)\\left(\\htmlData{anchor=pow-2}{-\\frac{1}{2}}\\right)(2+x)^{\\htmlData{anchor=new-pow-2}{-\\frac{3}{2}}} = -\\frac{1}{4}(2+x)^{-\\frac{3}{2}}$"},{"latex":"$$f''(0) = -\\frac{1}{4}(2)^{-\\frac{3}{2}} = -\\frac{1}{4}\\left(\\frac{1}{2\\sqrt{2}}\\right) = \\htmlData{anchor=f-double-prime-final}{-\\frac{\\sqrt{2}}{16}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"pow-2"},{"color":"green","style":"box","anchor":"f-double-prime-final"}],"connections":[{"to":"new-pow-2","from":"pow-2","color":"amber","label":"Power rule","style":"solid"}]},"order":3,"title":"Evaluate the second derivative","description":"Differentiating once more gives the second derivative."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f'''(x) = \\left(-\\frac{1}{4}\\right)\\left(\\htmlData{anchor=pow-3}{-\\frac{3}{2}}\\right)(2+x)^{\\htmlData{anchor=new-pow-3}{-\\frac{5}{2}}} = \\frac{3}{8}(2+x)^{-\\frac{5}{2}}$"},{"latex":"$$f'''(0) = \\frac{3}{8}(2)^{-\\frac{5}{2}} = \\frac{3}{8}\\left(\\frac{1}{4\\sqrt{2}}\\right) = \\htmlData{anchor=f-triple-prime-final}{\\frac{3\\sqrt{2}}{64}}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"pow-3"},{"color":"green","style":"box","anchor":"f-triple-prime-final"}],"connections":[{"to":"new-pow-3","from":"pow-3","color":"purple","label":"Power rule","style":"solid"}]},"order":4,"title":"Evaluate the third derivative","description":"Finally, we calculate the third derivative and its value at $x = 0$."},{"type":"text","order":5,"title":"Construct the Maclaurin series","content":"Now we substitute our values back into the Maclaurin series formula:\n\n$$f(x) \\approx f(0) + f'(0)x + \\frac{f''(0)}{2!}x^2 + \\frac{f'''(0)}{3!}x^3$$\n\nSubstituting the values we calculated:\n\n$$f(x) \\approx \\sqrt{2} + \\left(\\frac{\\sqrt{2}}{4}\\right)x + \\frac{-\\frac{\\sqrt{2}}{16}}{2}x^2 + \\frac{\\frac{3\\sqrt{2}}{64}}{6}x^3$$\n\nSimplifying each coefficient:\n\n- **For the $x^2$ term:**\n $$\\frac{-\\frac{\\sqrt{2}}{16}}{2} = -\\frac{\\sqrt{2}}{32}$$\n\n- **For the $x^3$ term:**\n $$\\frac{\\frac{3\\sqrt{2}}{64}}{6} = \\frac{\\sqrt{2}}{128}$$\n\nThis gives our final series expansion:\n\n$$\\sqrt{2} + \\frac{\\sqrt{2}}{4}x - \\frac{\\sqrt{2}}{32}x^2 + \\frac{\\sqrt{2}}{128}x^3$$"},{"type":"text","order":6,"title":"Determine the validity range","content":"For Part 2, the range of convergence of the series is based on the distance to the nearest singularity in the complex plane (at $x = -2$), or equivalently, using the binomial series condition $|y| < 1$ where $y = \\frac{x}{2}$:\n\n$$\\left|\\frac{x}{2}\\right| < 1 \\implies |x| < 2$$\n\nThis completes Part 2 and earns (**B1**)."},{"type":"text","order":7,"title":"Relate substitution to root","content":"For Part 3, we substitute $x = 0.5$ into our function $f(x) = (2+x)^{\\frac{1}{2}}$:\n\n$$f(0.5) = (2 + 0.5)^{\\frac{1}{2}} = \\sqrt{2.5} = \\sqrt{\\frac{5}{2}} = \\frac{\\sqrt{5}}{\\sqrt{2}} = \\frac{\\sqrt{10}}{2}$$\n\nThus, evaluating the series at $x = 0.5$ provides an approximation for $\\frac{\\sqrt{10}}{2}$ (**M1**)."},{"type":"text","order":8,"title":"Approximate the square root","content":"Substituting $x = 0.5$ into the series expansion we obtained:\n\n$$\\frac{\\sqrt{10}}{2} \\approx \\sqrt{2} + \\frac{\\sqrt{2}}{4}(0.5) - \\frac{\\sqrt{2}}{32}(0.5)^2 + \\frac{\\sqrt{2}}{128}(0.5)^3 \\approx 1.58132...\\text{ (**M1**)}$$\n\nMultiplying by $2$ to isolate $\\sqrt{10}$:\n\n$$\\sqrt{10} \\approx 2 \\times 1.58132... = 3.16264...$$\n\nRounding to 3 decimal places:\n\n$$\\sqrt{10} \\approx 3.163\\text{ (**A1**)}$$"}]},{"id":"75cc722a-d82e-4195-9a67-d1d5cd5b4f57:absolute_value_inequality","label":"(b) Absolute Value Inequality","steps":[{"type":"text","order":0,"title":"Recall the validity condition","content":"For a binomial expansion of the form $(1 + u)^n$ where $n$ is a non-integer, the expansion is valid (converges) if and only if the absolute value of the variable term, $u$, is strictly less than $1$: $$|u| < 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u = \\htmlData{anchor=u-term}{\\color{@sky}{\\frac{1}{2}x}}$"},{"latex":"$$\\left|\\htmlData{anchor=u-ineq}{\\color{@sky}{u}}\\right| < 1$"},{"latex":"$$\\left|\\htmlData{anchor=sub-ineq}{\\color{@sky}{\\frac{1}{2}x}}\\right| < 1$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"sky","style":"text-color","anchor":"sub-ineq"}],"connections":[{"to":"sub-ineq","from":"u-term","color":"sky","label":"substitute","style":"solid"},{"to":"sub-ineq","from":"u-ineq","color":"sky","label":null,"style":"dashed"}]},"order":1,"title":"Substitute the variable term","description":"In this expansion, our variable term is $u = \\frac{1}{2}x$. We set up the absolute value inequality using this term."},{"type":"text","order":2,"title":"Solve the inequality","content":"To find the range of values for $x$, we solve the inequality: $$\\left|\\frac{1}{2}x\\right| < 1$$ $$\\frac{1}{2}|x| < 1$$ $$|x| < 2$$ This can also be written as $-2 < x < 2$. Both formats represent the range of values for which the expansion is valid."}]},{"id":"75cc722a-d82e-4195-9a67-d1d5cd5b4f57:interval_inequality","label":"(b) Interval Inequality","steps":[{"type":"text","order":0,"title":"Recall the validity condition","content":"For a binomial expansion of $(1 + y)^n$ where $n$ is a fraction or negative, the expansion is only valid when the variable term $y$ lies strictly between $-1$ and $1$. In our case, the variable term in the bracket is $\\frac{1}{2}x$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Identify the term: } \\htmlData{anchor=term}{\\frac{1}{2}x}$"},{"latex":"$$-1 < \\htmlData{anchor=ineq}{\\frac{1}{2}x} < 1$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"term"},{"color":"sky","style":"text-color","anchor":"ineq"}],"connections":[{"to":"ineq","from":"term","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Set up the inequality","description":"We substitute our term, $\\frac{1}{2}x$, into the standard validity double inequality $-1 < y < 1$."},{"type":"text","order":2,"title":"Solve the double inequality","content":"To isolate $x$, we multiply all parts of the double inequality by $2$: $$-1 \\times 2 < x < 1 \\times 2$$ $$-2 < x < 2$$ This gives us the open interval $-2 < x < 2$. This can also be expressed in modulus notation as $|x| < 2$. Either form is fully correct and earns the B1 mark."}]},{"id":"4417af3b-ab13-4538-87dc-3673682081c8:direct_substitution","label":"(c) Direct Substitution and Algebraic Manipulation","steps":[{"type":"text","order":0,"title":"Relate $(2.5)^{\\frac{1}{2}}$ to $\\sqrt{10}$","content":"First, we substitute $x = 0.5$ into the original expression, $(2 + x)^{\\frac{1}{2}}$. This gives:\n\n$$(2 + 0.5)^{\\frac{1}{2}} = (2.5)^{\\frac{1}{2}} = \\sqrt{\\frac{5}{2}}$$\n\nTo relate this to $\\sqrt{10}$, we can rationalize the denominator by multiplying the numerator and denominator inside the square root by $2$:\n\n$$\\sqrt{\\frac{5}{2}} = \\sqrt{\\frac{5 \\times 2}{2 \\times 2}} = \\sqrt{\\frac{10}{4}} = \\frac{\\sqrt{10}}{2}$$\n\nThis shows us that:\n\n$$\\frac{\\sqrt{10}}{2} = (2 + 0.5)^{\\frac{1}{2}}$$"},{"type":"text","order":1,"title":"Evaluate the expansion","content":"Next, we substitute $x = 0.5$ into the binomial expansion of $(2 + x)^{\\frac{1}{2}}$ that was found in the previous part of the question. \n\nEvaluating the series expansion up to the given terms at $x = 0.5$ yields:\n\n$$(2 + 0.5)^{\\frac{1}{2}} \\approx 1.58132...$$\n\nSince we established that $(2 + 0.5)^{\\frac{1}{2}} = \\frac{\\sqrt{10}}{2}$, we can write:\n\n$$\\frac{\\sqrt{10}}{2} \\approx 1.58132...$$"},{"type":"text","order":2,"title":"Solve for $\\sqrt{10}$","content":"To find the approximation for $\\sqrt{10}$, we multiply both sides of our relationship by $2$:\n\n$$\\sqrt{10} \\approx 2 \\times 1.58132...$$\n\n$$\\sqrt{10} \\approx 3.16264...$$\n\nRounding this to $3$ decimal places, we get:\n\n$$\\sqrt{10} \\approx 3.163$$"}]}],"generatedAt":"2026-06-09T04:39:19.797Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":48},"walkthroughVideo":null},{"id":"4af758f8-ac50-4194-a926-01b49e3830c4","revisionId":"be6425ac-4d5a-44ca-ab54-f8903b2c9c00","questionNumber":null,"questionType":"LA","specification":"A geometric series is $a + ar + ar^2 + \\dots$","difficulty":2,"difficultyBand":"medium","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"214c1cdf-40d7-4b11-a954-1cc80030b106","label":"a","content":"Prove that the sum of the first $n$ terms of the series is \n\n$$S_n = \\frac{a(r^n - 1)}{r - 1}$$","markscheme":"- Write $S_n = a + ar + ar^2 + \\dots + ar^{n-1}$ M1\n- Write $rS_n = ar + ar^2 + \\dots + ar^n$ M1\n- Subtract equations: $rS_n - S_n = ar^n - a$ M1\n- Factorise and divide: $S_n(r - 1) = a(r^n - 1) \\implies S_n = \\frac{a(r^n - 1)}{r - 1}$ A1","marks":4,"order":0,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Write out the expression for $S_n$ up to the $(n-1)$-th term.","code":"M","acceptable":["S_n = a + ar + ar^2 + \\dots + ar^{n-1}"]},{"id":"p2","marks":1,"description":"Write out the expression for $rS_n$ by multiplying each term by $r$.","code":"M","acceptable":["rS_n = ar + ar^2 + ar^3 + \\dots + ar^n"]},{"id":"p3","marks":1,"description":"Subtract $S_n$ from $rS_n$ (or vice versa) to eliminate intermediate terms.","code":"M","acceptable":["rS_n - S_n = ar^n - a"]},{"id":"p4","marks":1,"description":"Factorise $S_n$ and $a$, and divide to achieve the target formula.","code":"A"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:05:32.294Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"algebraic_subtraction","label":"Algebraic Subtraction Method","steps":[{"type":"text","order":0,"title":"Set up the sum","content":"We begin by writing the sum of the first $n$ terms of a geometric series, which we denote as $S_n$.\n\nThe first term is $a$, the second is $ar$, and the $n$-th term is $ar^{n-1}$:\n\n$$S_n = a + ar + ar^2 + \\dots + ar^{n-1}$$\n\nWriting this equation down correctly earns the first mark."},{"type":"text","order":1,"title":"Multiply by the ratio","content":"Next, we multiply the entire equation by the common ratio $r$:\n\n$$rS_n = r(a + ar + ar^2 + \\dots + ar^{n-1})$$\n\nDistributing the $r$ across each term increases the power of $r$ by $1$ for every term:\n\n$$rS_n = ar + ar^2 + ar^3 + \\dots + ar^n$$\n\nThis step forms our second key equation and earns the second mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S_n = \\htmlData{anchor=term-a}{\\color{@pink}{a}} + \\htmlData{anchor=cancel-top}{\\color{@sky}{ar + ar^2 + \\dots + ar^{n-1}}}$"},{"latex":"$$rS_n = \\htmlData{anchor=cancel-bottom}{\\color{@sky}{ar + ar^2 + \\dots + ar^{n-1}}} + \\htmlData{anchor=term-arn}{\\color{@purple}{ar^n}}$"},{"latex":"$$rS_n - S_n = \\htmlData{anchor=diff-arn}{\\color{@purple}{ar^n}} - \\htmlData{anchor=diff-a}{\\color{@pink}{a}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"cancel-top"},{"color":"sky","style":"text-color","anchor":"cancel-bottom"}],"connections":[{"to":"cancel-bottom","from":"cancel-top","color":"sky","label":"cancels","style":"dashed"},{"to":"diff-a","from":"term-a","color":"pink","label":null,"style":"solid"},{"to":"diff-arn","from":"term-arn","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Subtract the equations","description":"Now, we subtract our first equation from our second equation. Notice how almost all terms in the middle cancel out, leaving only the first and last terms."},{"type":"text","order":3,"title":"Factorise and solve","content":"Now we factorise both sides of our equation to make $S_n$ the subject:\n\n- On the left-hand side, factor out $S_n$:\n $$S_n(r - 1)$$\n- On the right-hand side, factor out $a$:\n $$a(r^n - 1)$$\n\nThis gives us:\n\n$$S_n(r - 1) = a(r^n - 1)$$\n\nTo finish the proof, divide both sides by $(r - 1)$:\n\n$$S_n = \\frac{a(r^n - 1)}{r - 1}$$\n\nThis completes the proof and secures the final method and accuracy marks!"}]},{"id":"mathematical_induction","label":"Proof by Mathematical Induction","steps":[{"type":"text","order":0,"title":"Define the proposition","content":"To prove the formula for the sum of a geometric series using mathematical induction, we define our statement $P(n)$ for $n \\in \\mathbb{Z}^+$:\n\n$$S_n = a + ar + ar^2 + \\dots + ar^{n-1} = \\frac{a(r^n - 1)}{r - 1}$$\n\nwhere $a$ is the first term, $r$ is the common ratio ($r \\neq 1$), and $S_n$ is the sum of the first $n$ terms."},{"type":"text","order":1,"title":"Establish the base case","content":"We begin by testing the base case, which is $n = 1$.\n\nOn the left-hand side, the sum of the first $1$ term is simply the first term of the series:\n$$\\text{LHS} = S_1 = a$$\n\nOn the right-hand side, substituting $n = 1$ into the formula gives:\n$$\\text{RHS} = \\frac{a(r^1 - 1)}{r - 1} = \\frac{a(r - 1)}{r - 1} = a$$\n\nSince $\\text{LHS} = \\text{RHS}$, the proposition $P(1)$ is true."},{"type":"text","order":2,"title":"State the induction hypothesis","content":"We assume that the proposition is true for some positive integer $n = k$.\n\nThat is, we assume $P(k)$ is true:\n\n$$S_k = a + ar + ar^2 + \\dots + ar^{k-1} = \\frac{a(r^k - 1)}{r - 1}$$"},{"type":"text","order":3,"title":"Analyze the next term","content":"We now want to show that if $P(k)$ is true, then $P(k+1)$ is also true.\n\nThe sum of the first $k+1$ terms, $S_{k+1}$, is the sum of the first $k$ terms plus the $(k+1)^{\\text{th}}$ term:\n\n$$S_{k+1} = S_k + a r^k$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S_{k+1} = \\htmlData{anchor=sk-val}{S_k} + \\htmlData{anchor=term-add}{a r^k}$"},{"latex":"$$\\phantom{S_{k+1}} = \\htmlData{anchor=sk-sub}{\\frac{a(r^k - 1)}{r - 1}} + a r^k$"},{"latex":"$$S_{k+1} = \\frac{a(r^k - 1) + a r^k(r - 1)}{r - 1}$"},{"latex":"$$S_{k+1} = \\frac{a r^{k+1} - a}{r - 1} = \\frac{a(r^{k+1} - 1)}{r - 1}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"sk-sub"},{"color":"amber","style":"underline","anchor":"term-add"}],"connections":[{"to":"sk-sub","from":"sk-val","color":"sky","label":"Substitute S_k","style":"solid"}]},"order":4,"title":"Substitute and simplify","description":"We substitute our induction hypothesis expression for $S_k$ into the equation and find a common denominator to combine the terms."},{"type":"text","order":5,"title":"Write the conclusion","content":"We have shown that if $P(k)$ is true, then:\n\n$$S_{k+1} = \\frac{a(r^{k+1} - 1)}{r - 1}$$\n\nwhich is the proposition $P(k+1)$.\n\nSince the base case $P(1)$ is true, and $P(k) \\implies P(k+1)$, the formula $S_n = \\frac{a(r^n - 1)}{r - 1}$ is true for all $n \\in \\mathbb{Z}^+$ by mathematical induction.\n\n*(Note: Although mathematical induction is a rigorous proof method, standard exam markschemes for this specific syllabus question usually expect the direct algebraic proof of multiplying $S_n$ by $r$ and subtracting the two series equations to earn the full marks.)*"}]},{"id":"algebraic_identity","label":"Using the Difference of Powers Identity","steps":[{"type":"text","order":0,"title":"Define the series sum","content":"The sum of the first $n$ terms of a geometric series is represented by $S_n$. We write out the sum of these terms, from the first term $a$ up to the $n$-th term $ar^{n-1}$:\n\n$$S_n = a + ar + ar^2 + \\dots + ar^{n-1}$$\n\nThis expression earns you the first mark **(M1)** for correctly writing down the expansion of $S_n$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S_n = \\htmlData{anchor=s-unfactored}{a + ar + ar^2 + \\dots + ar^{n-1}}$"},{"latex":"$$S_n = \\htmlData{anchor=a-factor}{\\color{@sky}{a}}\\htmlData{anchor=s-factored}{\\left(1 + r + r^2 + \\dots + r^{n-1}\\right)}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-factor"}],"connections":[{"to":"s-factored","from":"s-unfactored","color":"sky","label":"factor out a","style":"solid"}]},"order":1,"title":"Factor out the first term","description":"To prepare our expression for the identity, we factor out the first term $a$ from every term in the sum."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=diff-identity}{\\color{@purple}{r^n - 1}} = (r - 1)\\htmlData{anchor=geom-part}{\\color{@amber}{\\left(1 + r + r^2 + \\dots + r^{n-1}\\right)}}$"},{"latex":"$$\\htmlData{anchor=geom-isolated}{\\color{@amber}{1 + r + r^2 + \\dots + r^{n-1}}} = \\frac{\\htmlData{anchor=diff-numerator}{\\color{@purple}{r^n - 1}}}{r - 1}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"diff-identity"},{"color":"amber","style":"text-color","anchor":"geom-part"}],"connections":[{"to":"geom-isolated","from":"geom-part","color":"amber","label":"isolate sum","style":"solid"}]},"order":2,"title":"Recall the difference of powers","description":"We now use the standard algebraic factorization identity for the difference of two $n$-th powers (specifically $r^n - 1$). For $r \\neq 1$, dividing both sides by $(r - 1)$ gives an explicit expression for the geometric progression factor."},{"type":"text","order":3,"title":"Substitute and complete proof","content":"Now we substitute this isolated sum back into our factored equation for $S_n$:\n\n$$S_n = a \\left( \\frac{r^n - 1}{r - 1} \\right)$$\n\n$$S_n = \\frac{a(r^n - 1)}{r - 1}$$\n\nThis algebraic manipulation earns the remaining marks **(M1, M1, A1)** as it is equivalent to factorising the difference of terms $rS_n - S_n$ and solving for $S_n$."}]}],"generatedAt":"2026-06-09T04:57:12.248Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"214c1cdf-40d7-4b11-a954-1cc80030b106:algebraic_subtraction","label":"Algebraic Subtraction Method","steps":[{"type":"text","order":0,"title":"Set up the sum","content":"We begin by writing the sum of the first $n$ terms of a geometric series, which we denote as $S_n$.\n\nThe first term is $a$, the second is $ar$, and the $n$-th term is $ar^{n-1}$:\n\n$$S_n = a + ar + ar^2 + \\dots + ar^{n-1}$$\n\nWriting this equation down correctly earns the first mark."},{"type":"text","order":1,"title":"Multiply by the ratio","content":"Next, we multiply the entire equation by the common ratio $r$:\n\n$$rS_n = r(a + ar + ar^2 + \\dots + ar^{n-1})$$\n\nDistributing the $r$ across each term increases the power of $r$ by $1$ for every term:\n\n$$rS_n = ar + ar^2 + ar^3 + \\dots + ar^n$$\n\nThis step forms our second key equation and earns the second mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S_n = \\htmlData{anchor=term-a}{\\color{@pink}{a}} + \\htmlData{anchor=cancel-top}{\\color{@sky}{ar + ar^2 + \\dots + ar^{n-1}}}$"},{"latex":"$$rS_n = \\htmlData{anchor=cancel-bottom}{\\color{@sky}{ar + ar^2 + \\dots + ar^{n-1}}} + \\htmlData{anchor=term-arn}{\\color{@purple}{ar^n}}$"},{"latex":"$$rS_n - S_n = \\htmlData{anchor=diff-arn}{\\color{@purple}{ar^n}} - \\htmlData{anchor=diff-a}{\\color{@pink}{a}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"cancel-top"},{"color":"sky","style":"text-color","anchor":"cancel-bottom"}],"connections":[{"to":"cancel-bottom","from":"cancel-top","color":"sky","label":"cancels","style":"dashed"},{"to":"diff-a","from":"term-a","color":"pink","label":null,"style":"solid"},{"to":"diff-arn","from":"term-arn","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Subtract the equations","description":"Now, we subtract our first equation from our second equation. Notice how almost all terms in the middle cancel out, leaving only the first and last terms."},{"type":"text","order":3,"title":"Factorise and solve","content":"Now we factorise both sides of our equation to make $S_n$ the subject:\n\n- On the left-hand side, factor out $S_n$:\n $$S_n(r - 1)$$\n- On the right-hand side, factor out $a$:\n $$a(r^n - 1)$$\n\nThis gives us:\n\n$$S_n(r - 1) = a(r^n - 1)$$\n\nTo finish the proof, divide both sides by $(r - 1)$:\n\n$$S_n = \\frac{a(r^n - 1)}{r - 1}$$\n\nThis completes the proof and secures the final method and accuracy marks!"}]},{"id":"214c1cdf-40d7-4b11-a954-1cc80030b106:mathematical_induction","label":"Proof by Mathematical Induction","steps":[{"type":"text","order":0,"title":"Define the proposition","content":"To prove the formula for the sum of a geometric series using mathematical induction, we define our statement $P(n)$ for $n \\in \\mathbb{Z}^+$:\n\n$$S_n = a + ar + ar^2 + \\dots + ar^{n-1} = \\frac{a(r^n - 1)}{r - 1}$$\n\nwhere $a$ is the first term, $r$ is the common ratio ($r \\neq 1$), and $S_n$ is the sum of the first $n$ terms."},{"type":"text","order":1,"title":"Establish the base case","content":"We begin by testing the base case, which is $n = 1$.\n\nOn the left-hand side, the sum of the first $1$ term is simply the first term of the series:\n$$\\text{LHS} = S_1 = a$$\n\nOn the right-hand side, substituting $n = 1$ into the formula gives:\n$$\\text{RHS} = \\frac{a(r^1 - 1)}{r - 1} = \\frac{a(r - 1)}{r - 1} = a$$\n\nSince $\\text{LHS} = \\text{RHS}$, the proposition $P(1)$ is true."},{"type":"text","order":2,"title":"State the induction hypothesis","content":"We assume that the proposition is true for some positive integer $n = k$.\n\nThat is, we assume $P(k)$ is true:\n\n$$S_k = a + ar + ar^2 + \\dots + ar^{k-1} = \\frac{a(r^k - 1)}{r - 1}$$"},{"type":"text","order":3,"title":"Analyze the next term","content":"We now want to show that if $P(k)$ is true, then $P(k+1)$ is also true.\n\nThe sum of the first $k+1$ terms, $S_{k+1}$, is the sum of the first $k$ terms plus the $(k+1)^{\\text{th}}$ term:\n\n$$S_{k+1} = S_k + a r^k$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S_{k+1} = \\htmlData{anchor=sk-val}{S_k} + \\htmlData{anchor=term-add}{a r^k}$"},{"latex":"$$\\phantom{S_{k+1}} = \\htmlData{anchor=sk-sub}{\\frac{a(r^k - 1)}{r - 1}} + a r^k$"},{"latex":"$$S_{k+1} = \\frac{a(r^k - 1) + a r^k(r - 1)}{r - 1}$"},{"latex":"$$S_{k+1} = \\frac{a r^{k+1} - a}{r - 1} = \\frac{a(r^{k+1} - 1)}{r - 1}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"sk-sub"},{"color":"amber","style":"underline","anchor":"term-add"}],"connections":[{"to":"sk-sub","from":"sk-val","color":"sky","label":"Substitute S_k","style":"solid"}]},"order":4,"title":"Substitute and simplify","description":"We substitute our induction hypothesis expression for $S_k$ into the equation and find a common denominator to combine the terms."},{"type":"text","order":5,"title":"Write the conclusion","content":"We have shown that if $P(k)$ is true, then:\n\n$$S_{k+1} = \\frac{a(r^{k+1} - 1)}{r - 1}$$\n\nwhich is the proposition $P(k+1)$.\n\nSince the base case $P(1)$ is true, and $P(k) \\implies P(k+1)$, the formula $S_n = \\frac{a(r^n - 1)}{r - 1}$ is true for all $n \\in \\mathbb{Z}^+$ by mathematical induction.\n\n*(Note: Although mathematical induction is a rigorous proof method, standard exam markschemes for this specific syllabus question usually expect the direct algebraic proof of multiplying $S_n$ by $r$ and subtracting the two series equations to earn the full marks.)*"}]},{"id":"214c1cdf-40d7-4b11-a954-1cc80030b106:algebraic_identity","label":"Using the Difference of Powers Identity","steps":[{"type":"text","order":0,"title":"Define the series sum","content":"The sum of the first $n$ terms of a geometric series is represented by $S_n$. We write out the sum of these terms, from the first term $a$ up to the $n$-th term $ar^{n-1}$:\n\n$$S_n = a + ar + ar^2 + \\dots + ar^{n-1}$$\n\nThis expression earns you the first mark **(M1)** for correctly writing down the expansion of $S_n$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S_n = \\htmlData{anchor=s-unfactored}{a + ar + ar^2 + \\dots + ar^{n-1}}$"},{"latex":"$$S_n = \\htmlData{anchor=a-factor}{\\color{@sky}{a}}\\htmlData{anchor=s-factored}{\\left(1 + r + r^2 + \\dots + r^{n-1}\\right)}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-factor"}],"connections":[{"to":"s-factored","from":"s-unfactored","color":"sky","label":"factor out a","style":"solid"}]},"order":1,"title":"Factor out the first term","description":"To prepare our expression for the identity, we factor out the first term $a$ from every term in the sum."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=diff-identity}{\\color{@purple}{r^n - 1}} = (r - 1)\\htmlData{anchor=geom-part}{\\color{@amber}{\\left(1 + r + r^2 + \\dots + r^{n-1}\\right)}}$"},{"latex":"$$\\htmlData{anchor=geom-isolated}{\\color{@amber}{1 + r + r^2 + \\dots + r^{n-1}}} = \\frac{\\htmlData{anchor=diff-numerator}{\\color{@purple}{r^n - 1}}}{r - 1}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"diff-identity"},{"color":"amber","style":"text-color","anchor":"geom-part"}],"connections":[{"to":"geom-isolated","from":"geom-part","color":"amber","label":"isolate sum","style":"solid"}]},"order":2,"title":"Recall the difference of powers","description":"We now use the standard algebraic factorization identity for the difference of two $n$-th powers (specifically $r^n - 1$). For $r \\neq 1$, dividing both sides by $(r - 1)$ gives an explicit expression for the geometric progression factor."},{"type":"text","order":3,"title":"Substitute and complete proof","content":"Now we substitute this isolated sum back into our factored equation for $S_n$:\n\n$$S_n = a \\left( \\frac{r^n - 1}{r - 1} \\right)$$\n\n$$S_n = \\frac{a(r^n - 1)}{r - 1}$$\n\nThis algebraic manipulation earns the remaining marks **(M1, M1, A1)** as it is equivalent to factorising the difference of terms $rS_n - S_n$ and solving for $S_n$."}]}],"generatedAt":"2026-06-09T04:57:23.759Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":55},"walkthroughVideo":null},{"id":"567a339e-ed5f-49fa-b1f0-ea25bac54256","revisionId":"1cab7c1b-e8a3-427f-9aad-edd3a91889c5","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":"medium","tier":null,"totalMarks":7,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"c3ed493f-0b35-4394-a677-9e3ad10ef093","label":"a","content":"Expand $\\sqrt[4]{1+4x}$ in ascending powers of $x$ up to and including $x^3$ and state the set of values of $x$ for which the expansion is valid.","markscheme":"- Writes as $(1+4x)^{\\frac{1}{4}}$ and applies expansion: $1 + (\\frac{1}{4})(4x) + \\frac{(\\frac{1}{4})(-\\frac{3}{4})}{2}(4x)^2 + \\frac{(\\frac{1}{4})(-\\frac{3}{4})(-\\frac{7}{4})}{6}(4x)^3$ M1\n- Correct $x$ term: $x$ A1\n- Correct $x^2$ term: $-\\frac{3}{2}x^2$ A1\n- Correct $x^3$ term: $\\frac{7}{2}x^3$ A1\n- Validity: $|4x| < 1 \\Rightarrow |x| < \\frac{1}{4}$ B1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Evidence of binomial expansion formula for $(1+ax)^n$ with $n=1/4$ and $a=4$.","code":"M"},{"id":"p2","marks":1,"description":"Correct simplified term in $x$.","code":"A"},{"id":"p3","marks":1,"description":"Correct simplified term in $x^2$.","code":"A"},{"id":"p4","marks":1,"description":"Correct simplified term in $x^3$.","code":"A"},{"id":"p5","marks":1,"description":"Correct validity range for $|x|$.","code":"B"}],"finalAnswers":["1 + x - \\frac{3}{2}x^2 + \\frac{7}{2}x^3","|x| < \\frac{1}{4}"],"generatedAt":"2026-05-06T14:03:51.631Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"binomial_expansion","label":"Binomial Series Formula","steps":[{"type":"text","order":0,"title":"Convert to index form","content":"To expand $\\sqrt[4]{1+4x}$, we first write the radical expression in index form as a power:\n\n$$\\sqrt[4]{1+4x} = (1+4x)^{\\frac{1}{4}}$$\n\nThis fits the general binomial form $(1+y)^n$ where $n = \\frac{1}{4}$ and $y = 4x$."},{"type":"text","order":1,"title":"State binomial formula","content":"Recall the general binomial expansion formula for any real number $n$:\n\n$$(1+y)^n = 1 + ny + \\frac{n(n-1)}{2!}y^2 + \\frac{n(n-1)(n-2)}{3!}y^3 + \\dots$$\n\nHere, we substitute $n = \\frac{1}{4}$ and $y = 4x$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=y-val}{\\color{@sky}{4x}}$"},{"latex":"$$n = \\htmlData{anchor=n-val}{\\color{@amber}{\\frac{1}{4}}}$"},{"latex":"$$1 + \\htmlData{anchor=n1}{\\color{@amber}{n}}\\htmlData{anchor=y1}{\\color{@sky}{y}} + \\frac{\\htmlData{anchor=n2}{\\color{@amber}{n}}(\\htmlData{anchor=n3}{\\color{@amber}{n}}-1)}{2}\\htmlData{anchor=y2}{\\color{@sky}{y}}^2 + \\frac{\\htmlData{anchor=n4}{\\color{@amber}{n}}(\\htmlData{anchor=n5}{\\color{@amber}{n}}-1)(\\htmlData{anchor=n6}{\\color{@amber}{n}}-2)}{6}\\htmlData{anchor=y3}{\\color{@sky}{y}}^3$"},{"latex":"$$= 1 + \\color{@amber}{\\left(\\frac{1}{4}\\right)}\\color{@sky}{(4x)} + \\frac{\\color{@amber}{\\left(\\frac{1}{4}\\right)}\\color{@amber}{\\left(-\\frac{3}{4}\\right)}}{2}\\color{@sky}{(4x)^2} + \\frac{\\color{@amber}{\\left(\\frac{1}{4}\\right)}\\color{@amber}{\\left(-\\frac{3}{4}\\right)}\\color{@amber}{\\left(-\\frac{7}{4}\\right)}}{6}\\color{@sky}{(4x)^3}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-val"},{"color":"amber","style":"text-color","anchor":"n-val"}],"connections":[{"to":"y1","from":"y-val","color":"sky","label":null,"style":"solid"},{"to":"n1","from":"n-val","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Substitute into formula","description":"We substitute $n = \\frac{1}{4}$ and $y = 4x$ into the first four terms of the binomial expansion."},{"type":"text","order":3,"title":"Simplify terms","content":"Now we simplify each term individually:\n\n* **First-degree term ($x$):**\n $$\\left(\\frac{1}{4}\\right)(4x) = x$$\n\n* **Second-degree term ($x^2$):**\n $$\\frac{\\left(\\frac{1}{4}\\right)\\left(-\\frac{3}{4}\\right)}{2}(16x^2) = \\frac{-\\frac{3}{16}}{2}(16x^2) = -\\frac{3}{32}(16x^2) = -\\frac{3}{2}x^2$$\n\n* **Third-degree term ($x^3$):**\n $$\\frac{\\left(\\frac{1}{4}\\right)\\left(-\\frac{3}{4}\\right)\\left(-\\frac{7}{4}\\right)}{6}(64x^3) = \\frac{\\frac{21}{64}}{6}(64x^3) = \\frac{21}{384}(64x^3) = \\frac{7}{2}x^3$$\n\nPutting them all together, we get:\n\n$$\\sqrt[4]{1+4x} = 1 + x - \\frac{3}{2}x^2 + \\frac{7}{2}x^3 + \\dots$$"},{"type":"text","order":4,"title":"Find the validity","content":"The binomial series expansion $(1+y)^n$ is valid if and only if $|y| < 1$.\n\nSubstituting $y = 4x$:\n\n$$|4x| < 1 \\implies 4|x| < 1 \\implies |x| < \\frac{1}{4}$$\n\nTherefore, the set of values for which the expansion is valid is $|x| < \\frac{1}{4}$ (or $-\\frac{1}{4} < x < \\frac{1}{4}$)."},{"type":"text","order":5,"title":"Approximate the value","content":"To approximate $\\sqrt[4]{1.04}$ using our expansion, we find the value of $x$ by setting the term inside the radical equal to $1.04$:\n\n$$1 + 4x = 1.04 \\implies 4x = 0.04 \\implies x = 0.01$$\n\nSince $|0.01| < \\frac{1}{4}$, we can substitute $x = 0.01$ into our binomial expansion:\n\n$$\\sqrt[4]{1.04} \\approx 1 + (0.01) - \\frac{3}{2}(0.01)^2 + \\frac{7}{2}(0.01)^3$$\n$$= 1 + 0.01 - 1.5(0.0001) + 3.5(0.000001)$$\n$$= 1 + 0.01 - 0.00015 + 0.0000035$$\n$$= 1.0098535 \\approx 1.00985$$\n\nRounding to 5 decimal places gives $1.00985$."}]},{"id":"maclaurin_series","label":"Maclaurin Series Expansion","steps":[{"type":"text","order":0,"title":"Recall the Maclaurin Series","content":"The Maclaurin series expansion of a function $f(x)$ up to the $x^3$ term is given by:\n\n$$f(x) \\approx f(0) + f'(0)x + \\frac{f''(0)}{2!}x^2 + \\frac{f'''(0)}{3!}x^3$$\n\nLet's define our function as:\n\n$$f(x) = (1+4x)^{\\frac{1}{4}}$$\n\nTo find the series expansion, we need to calculate $f(0)$ and the values of the first three derivatives of $f(x)$ evaluated at $x = 0$."},{"type":"text","order":1,"title":"Calculate first derivative","content":"Using the chain rule, we differentiate $f(x) = (1+4x)^{\\frac{1}{4}}$:\n\n$$f'(x) = \\frac{1}{4}(1+4x)^{-\\frac{3}{4}} \\cdot 4 = (1+4x)^{-\\frac{3}{4}}$$\n\nNow, evaluate $f(0)$ and $f'(0)$ by substituting $x = 0$:\n\n$$f(0) = (1+0)^{\\frac{1}{4}} = 1$$\n$$f'(0) = (1+0)^{-\\frac{3}{4}} = 1$$"},{"type":"text","order":2,"title":"Calculate higher derivatives","content":"Next, we find the second and third derivatives of $f(x)$ by differentiating $f'(x) = (1+4x)^{-\\frac{3}{4}}$ repeatedly using the chain rule:\n\n$$f''(x) = -\\frac{3}{4}(1+4x)^{-\\frac{7}{4}} \\cdot 4 = -3(1+4x)^{-\\frac{7}{4}}$$\n$$f'''(x) = -3 \\left(-\\frac{7}{4}\\right)(1+4x)^{-\\frac{11}{4}} \\cdot 4 = 21(1+4x)^{-\\frac{11}{4}}$$\n\nNow, evaluate these derivatives at $x = 0$:\n\n$$f''(0) = -3(1+0)^{-\\frac{7}{4}} = -3$$\n$$f'''(0) = 21(1+0)^{-\\frac{11}{4}} = 21$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(x) \\approx \\htmlData{anchor=f0}{\\color{@sky}{1}} + \\htmlData{anchor=f1}{\\color{@amber}{1}}x + \\frac{\\htmlData{anchor=f2}{\\color{@purple}{-3}}}{2!}x^2 + \\frac{\\htmlData{anchor=f3}{\\color{@teal}{21}}}{3!}x^3$"},{"text":"Simplify the coefficients of each term:"},{"latex":"$$f(x) \\approx 1 + x - \\frac{3}{2}x^2 + \\frac{21}{6}x^3$"},{"latex":"$$f(x) \\approx 1 + x - \\frac{3}{2}x^2 + \\frac{7}{2}x^3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"f0"},{"color":"amber","style":"text-color","anchor":"f1"},{"color":"purple","style":"text-color","anchor":"f2"},{"color":"teal","style":"text-color","anchor":"f3"}],"connections":[]},"order":3,"title":"Substitute into Maclaurin formula","description":"We substitute our calculated values $f(0) = 1$, $f'(0) = 1$, $f''(0) = -3$, and $f'''(0) = 21$ into the Maclaurin series expansion formula."},{"type":"text","order":4,"title":"Find interval of validity","content":"For a standard binomial-type expansion $(1 + u)^n$ where $n$ is a non-integer, the series converges and is valid when $|u| < 1$.\n\nIn our function, the variable expression playing the role of $u$ is $4x$. Therefore, the expansion is valid when:\n\n$$|4x| < 1$$\n\nSolving for $x$, we divide by $4$:\n\n$$|x| < \\frac{1}{4}$$\n\nThis can also be written as $-\\frac{1}{4} < x < \\frac{1}{4}$."},{"type":"text","order":5,"title":"Approximate the value","content":"We can approximate $\\sqrt[4]{1.04}$ by substituting a suitable value of $x$ into our Maclaurin series expansion:\n\n$$1 + 4x = 1.04 \\implies 4x = 0.04 \\implies x = 0.01$$\n\nSubstituting $x = 0.01$ into the series:\n\n$$\\sqrt[4]{1.04} \\approx 1 + (0.01) - \\frac{3}{2}(0.01)^2 + \\frac{7}{2}(0.01)^3$$\n$$= 1 + 0.01 - 1.5(0.0001) + 3.5(0.000001)$$\n$$= 1 + 0.01 - 0.00015 + 0.0000035$$\n$$= 1.0098535 \\approx 1.00985$$\n\nRounding to 5 decimal places gives $1.00985$."}]}],"generatedAt":"2026-06-09T04:50:30.872Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"11c23a38-8d6f-4fe4-a324-ff4301b4279d","label":"b","content":"Use your expansion to find an approximation for $\\sqrt[4]{1.04}$ to 5 decimal places.","markscheme":"- Sets $1+4x = 1.04 \\Rightarrow x = 0.01$ M1\n- Substitutes $x=0.01$ into expansion: $1 + (0.01) - 1.5(0.01)^2 + 3.5(0.01)^3 \\approx 1.00985$ A1","marks":2,"order":1,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Identification of the correct value of $x$ for the substitution.","code":"M"},{"id":"p2","marks":1,"description":"Correct numerical approximation to 5 decimal places.","code":"A"}],"finalAnswers":["1.00985"],"generatedAt":"2026-05-06T14:03:51.631Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Algebraic Substitution","steps":[{"type":"text","order":0,"title":"Equate the terms","content":"To approximate $\\sqrt[4]{1.04}$ using our binomial expansion of $\\sqrt[4]{1+4x}$, we first equate the terms under the fourth-root radical to find the corresponding value of $x$.\n\nWe set the expression under the root, $1+4x$, equal to $1.04$:\n\n$$1 + 4x = 1.04$$","highlights":[{"text":"1+4x = 1.04","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Subtract 1 from both sides:"},{"latex":"$$4\\htmlData{anchor=x-var}{\\color{@sky}{x}} = 0.04$"},{"text":"Divide by 4:"},{"latex":"$$\\htmlData{anchor=x-val}{\\color{@sky}{x}} = 0.01$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-var"},{"color":"sky","style":"text-color","anchor":"x-val"}],"connections":[{"to":"x-val","from":"x-var","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Solve for x","description":"Solving this simple linear equation gives the substitution value for our expansion, earning the first method mark (M1)."},{"type":"text","order":2,"title":"Substitute into expansion","content":"Now, we substitute $x = 0.01$ directly into our expanded series:\n\n$$\\sqrt[4]{1+4x} \\approx 1 + x - 1.5x^2 + 3.5x^3$$"},{"type":"text","order":3,"title":"Evaluate the approximation","content":"Substituting $x = 0.01$ into the binomial series gives:\n\n$$\\sqrt[4]{1.04} \\approx 1 + (0.01) - 1.5(0.01)^2 + 3.5(0.01)^3$$\n\n$$= 1 + 0.01 - 1.5(0.0001) + 3.5(0.000001)$$\n\n$$= 1 + 0.01 - 0.00015 + 0.0000035 = 1.0098535$$\n\nRounding this value to $5$ decimal places gives:\n\n$$\\sqrt[4]{1.04} \\approx 1.00985$$\n\nThis completes the calculation and earns the final accuracy mark (A1).","calculator":[{"gdc":{"expression":"1 + 0.01 - 1.5 * 0.01^2 + 3.5 * 0.01^3","instructions":"Enter the following in your main calculation screen:\n1 + 0.01 - 1.5 * 0.01^2 + 3.5 * 0.01^3\nExecute the operation to observe the value."},"ti84":{"expression":"1 + 0.01 - 1.5 * 0.01^2 + 3.5 * 0.01^3","instructions":"Type the expression into the home screen:\n1 + 0.01 - 1.5 * 0.01^2 + 3.5 * 0.01^3\nPress ENTER to get the result."},"desmos":{"expression":"1 + 0.01 - 1.5 * 0.01^2 + 3.5 * 0.01^3","instructions":"Type the arithmetic expression directly into a cell on Desmos to calculate the value."},"description":"Calculate the final numerical approximation using a calculator to ensure high precision before rounding."}]}]}],"generatedAt":"2026-06-09T04:51:38.223Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"c3ed493f-0b35-4394-a677-9e3ad10ef093:binomial_expansion","label":"(a) Binomial Series Formula","steps":[{"type":"text","order":0,"title":"Convert to index form","content":"To expand $\\sqrt[4]{1+4x}$, we first write the radical expression in index form as a power:\n\n$$\\sqrt[4]{1+4x} = (1+4x)^{\\frac{1}{4}}$$\n\nThis fits the general binomial form $(1+y)^n$ where $n = \\frac{1}{4}$ and $y = 4x$."},{"type":"text","order":1,"title":"State binomial formula","content":"Recall the general binomial expansion formula for any real number $n$:\n\n$$(1+y)^n = 1 + ny + \\frac{n(n-1)}{2!}y^2 + \\frac{n(n-1)(n-2)}{3!}y^3 + \\dots$$\n\nHere, we substitute $n = \\frac{1}{4}$ and $y = 4x$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=y-val}{\\color{@sky}{4x}}$"},{"latex":"$$n = \\htmlData{anchor=n-val}{\\color{@amber}{\\frac{1}{4}}}$"},{"latex":"$$1 + \\htmlData{anchor=n1}{\\color{@amber}{n}}\\htmlData{anchor=y1}{\\color{@sky}{y}} + \\frac{\\htmlData{anchor=n2}{\\color{@amber}{n}}(\\htmlData{anchor=n3}{\\color{@amber}{n}}-1)}{2}\\htmlData{anchor=y2}{\\color{@sky}{y}}^2 + \\frac{\\htmlData{anchor=n4}{\\color{@amber}{n}}(\\htmlData{anchor=n5}{\\color{@amber}{n}}-1)(\\htmlData{anchor=n6}{\\color{@amber}{n}}-2)}{6}\\htmlData{anchor=y3}{\\color{@sky}{y}}^3$"},{"latex":"$$= 1 + \\color{@amber}{\\left(\\frac{1}{4}\\right)}\\color{@sky}{(4x)} + \\frac{\\color{@amber}{\\left(\\frac{1}{4}\\right)}\\color{@amber}{\\left(-\\frac{3}{4}\\right)}}{2}\\color{@sky}{(4x)^2} + \\frac{\\color{@amber}{\\left(\\frac{1}{4}\\right)}\\color{@amber}{\\left(-\\frac{3}{4}\\right)}\\color{@amber}{\\left(-\\frac{7}{4}\\right)}}{6}\\color{@sky}{(4x)^3}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-val"},{"color":"amber","style":"text-color","anchor":"n-val"}],"connections":[{"to":"y1","from":"y-val","color":"sky","label":null,"style":"solid"},{"to":"n1","from":"n-val","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Substitute into formula","description":"We substitute $n = \\frac{1}{4}$ and $y = 4x$ into the first four terms of the binomial expansion."},{"type":"text","order":3,"title":"Simplify terms","content":"Now we simplify each term individually:\n\n* **First-degree term ($x$):**\n $$\\left(\\frac{1}{4}\\right)(4x) = x$$\n\n* **Second-degree term ($x^2$):**\n $$\\frac{\\left(\\frac{1}{4}\\right)\\left(-\\frac{3}{4}\\right)}{2}(16x^2) = \\frac{-\\frac{3}{16}}{2}(16x^2) = -\\frac{3}{32}(16x^2) = -\\frac{3}{2}x^2$$\n\n* **Third-degree term ($x^3$):**\n $$\\frac{\\left(\\frac{1}{4}\\right)\\left(-\\frac{3}{4}\\right)\\left(-\\frac{7}{4}\\right)}{6}(64x^3) = \\frac{\\frac{21}{64}}{6}(64x^3) = \\frac{21}{384}(64x^3) = \\frac{7}{2}x^3$$\n\nPutting them all together, we get:\n\n$$\\sqrt[4]{1+4x} = 1 + x - \\frac{3}{2}x^2 + \\frac{7}{2}x^3 + \\dots$$"},{"type":"text","order":4,"title":"Find the validity","content":"The binomial series expansion $(1+y)^n$ is valid if and only if $|y| < 1$.\n\nSubstituting $y = 4x$:\n\n$$|4x| < 1 \\implies 4|x| < 1 \\implies |x| < \\frac{1}{4}$$\n\nTherefore, the set of values for which the expansion is valid is $|x| < \\frac{1}{4}$ (or $-\\frac{1}{4} < x < \\frac{1}{4}$)."},{"type":"text","order":5,"title":"Approximate the value","content":"To approximate $\\sqrt[4]{1.04}$ using our expansion, we find the value of $x$ by setting the term inside the radical equal to $1.04$:\n\n$$1 + 4x = 1.04 \\implies 4x = 0.04 \\implies x = 0.01$$\n\nSince $|0.01| < \\frac{1}{4}$, we can substitute $x = 0.01$ into our binomial expansion:\n\n$$\\sqrt[4]{1.04} \\approx 1 + (0.01) - \\frac{3}{2}(0.01)^2 + \\frac{7}{2}(0.01)^3$$\n$$= 1 + 0.01 - 1.5(0.0001) + 3.5(0.000001)$$\n$$= 1 + 0.01 - 0.00015 + 0.0000035$$\n$$= 1.0098535 \\approx 1.00985$$\n\nRounding to 5 decimal places gives $1.00985$."}]},{"id":"c3ed493f-0b35-4394-a677-9e3ad10ef093:maclaurin_series","label":"(a) Maclaurin Series Expansion","steps":[{"type":"text","order":0,"title":"Recall the Maclaurin Series","content":"The Maclaurin series expansion of a function $f(x)$ up to the $x^3$ term is given by:\n\n$$f(x) \\approx f(0) + f'(0)x + \\frac{f''(0)}{2!}x^2 + \\frac{f'''(0)}{3!}x^3$$\n\nLet's define our function as:\n\n$$f(x) = (1+4x)^{\\frac{1}{4}}$$\n\nTo find the series expansion, we need to calculate $f(0)$ and the values of the first three derivatives of $f(x)$ evaluated at $x = 0$."},{"type":"text","order":1,"title":"Calculate first derivative","content":"Using the chain rule, we differentiate $f(x) = (1+4x)^{\\frac{1}{4}}$:\n\n$$f'(x) = \\frac{1}{4}(1+4x)^{-\\frac{3}{4}} \\cdot 4 = (1+4x)^{-\\frac{3}{4}}$$\n\nNow, evaluate $f(0)$ and $f'(0)$ by substituting $x = 0$:\n\n$$f(0) = (1+0)^{\\frac{1}{4}} = 1$$\n$$f'(0) = (1+0)^{-\\frac{3}{4}} = 1$$"},{"type":"text","order":2,"title":"Calculate higher derivatives","content":"Next, we find the second and third derivatives of $f(x)$ by differentiating $f'(x) = (1+4x)^{-\\frac{3}{4}}$ repeatedly using the chain rule:\n\n$$f''(x) = -\\frac{3}{4}(1+4x)^{-\\frac{7}{4}} \\cdot 4 = -3(1+4x)^{-\\frac{7}{4}}$$\n$$f'''(x) = -3 \\left(-\\frac{7}{4}\\right)(1+4x)^{-\\frac{11}{4}} \\cdot 4 = 21(1+4x)^{-\\frac{11}{4}}$$\n\nNow, evaluate these derivatives at $x = 0$:\n\n$$f''(0) = -3(1+0)^{-\\frac{7}{4}} = -3$$\n$$f'''(0) = 21(1+0)^{-\\frac{11}{4}} = 21$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(x) \\approx \\htmlData{anchor=f0}{\\color{@sky}{1}} + \\htmlData{anchor=f1}{\\color{@amber}{1}}x + \\frac{\\htmlData{anchor=f2}{\\color{@purple}{-3}}}{2!}x^2 + \\frac{\\htmlData{anchor=f3}{\\color{@teal}{21}}}{3!}x^3$"},{"text":"Simplify the coefficients of each term:"},{"latex":"$$f(x) \\approx 1 + x - \\frac{3}{2}x^2 + \\frac{21}{6}x^3$"},{"latex":"$$f(x) \\approx 1 + x - \\frac{3}{2}x^2 + \\frac{7}{2}x^3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"f0"},{"color":"amber","style":"text-color","anchor":"f1"},{"color":"purple","style":"text-color","anchor":"f2"},{"color":"teal","style":"text-color","anchor":"f3"}],"connections":[]},"order":3,"title":"Substitute into Maclaurin formula","description":"We substitute our calculated values $f(0) = 1$, $f'(0) = 1$, $f''(0) = -3$, and $f'''(0) = 21$ into the Maclaurin series expansion formula."},{"type":"text","order":4,"title":"Find interval of validity","content":"For a standard binomial-type expansion $(1 + u)^n$ where $n$ is a non-integer, the series converges and is valid when $|u| < 1$.\n\nIn our function, the variable expression playing the role of $u$ is $4x$. Therefore, the expansion is valid when:\n\n$$|4x| < 1$$\n\nSolving for $x$, we divide by $4$:\n\n$$|x| < \\frac{1}{4}$$\n\nThis can also be written as $-\\frac{1}{4} < x < \\frac{1}{4}$."},{"type":"text","order":5,"title":"Approximate the value","content":"We can approximate $\\sqrt[4]{1.04}$ by substituting a suitable value of $x$ into our Maclaurin series expansion:\n\n$$1 + 4x = 1.04 \\implies 4x = 0.04 \\implies x = 0.01$$\n\nSubstituting $x = 0.01$ into the series:\n\n$$\\sqrt[4]{1.04} \\approx 1 + (0.01) - \\frac{3}{2}(0.01)^2 + \\frac{7}{2}(0.01)^3$$\n$$= 1 + 0.01 - 1.5(0.0001) + 3.5(0.000001)$$\n$$= 1 + 0.01 - 0.00015 + 0.0000035$$\n$$= 1.0098535 \\approx 1.00985$$\n\nRounding to 5 decimal places gives $1.00985$."}]},{"id":"11c23a38-8d6f-4fe4-a324-ff4301b4279d:substitution","label":"(b) Algebraic Substitution","steps":[{"type":"text","order":0,"title":"Equate the terms","content":"To approximate $\\sqrt[4]{1.04}$ using our binomial expansion of $\\sqrt[4]{1+4x}$, we first equate the terms under the fourth-root radical to find the corresponding value of $x$.\n\nWe set the expression under the root, $1+4x$, equal to $1.04$:\n\n$$1 + 4x = 1.04$$","highlights":[{"text":"1+4x = 1.04","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Subtract 1 from both sides:"},{"latex":"$$4\\htmlData{anchor=x-var}{\\color{@sky}{x}} = 0.04$"},{"text":"Divide by 4:"},{"latex":"$$\\htmlData{anchor=x-val}{\\color{@sky}{x}} = 0.01$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-var"},{"color":"sky","style":"text-color","anchor":"x-val"}],"connections":[{"to":"x-val","from":"x-var","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Solve for x","description":"Solving this simple linear equation gives the substitution value for our expansion, earning the first method mark (M1)."},{"type":"text","order":2,"title":"Substitute into expansion","content":"Now, we substitute $x = 0.01$ directly into our expanded series:\n\n$$\\sqrt[4]{1+4x} \\approx 1 + x - 1.5x^2 + 3.5x^3$$"},{"type":"text","order":3,"title":"Evaluate the approximation","content":"Substituting $x = 0.01$ into the binomial series gives:\n\n$$\\sqrt[4]{1.04} \\approx 1 + (0.01) - 1.5(0.01)^2 + 3.5(0.01)^3$$\n\n$$= 1 + 0.01 - 1.5(0.0001) + 3.5(0.000001)$$\n\n$$= 1 + 0.01 - 0.00015 + 0.0000035 = 1.0098535$$\n\nRounding this value to $5$ decimal places gives:\n\n$$\\sqrt[4]{1.04} \\approx 1.00985$$\n\nThis completes the calculation and earns the final accuracy mark (A1).","calculator":[{"gdc":{"expression":"1 + 0.01 - 1.5 * 0.01^2 + 3.5 * 0.01^3","instructions":"Enter the following in your main calculation screen:\n1 + 0.01 - 1.5 * 0.01^2 + 3.5 * 0.01^3\nExecute the operation to observe the value."},"ti84":{"expression":"1 + 0.01 - 1.5 * 0.01^2 + 3.5 * 0.01^3","instructions":"Type the expression into the home screen:\n1 + 0.01 - 1.5 * 0.01^2 + 3.5 * 0.01^3\nPress ENTER to get the result."},"desmos":{"expression":"1 + 0.01 - 1.5 * 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3.4 Sequences and Series (A-level only)

Question 7

3.4 Sequences and Series (A-level only) Questions

3.4 Sequences and Series (A-level only)

Question 7

3.4 Sequences and Series (A-level only) Questions