de the parentheses equal:\n\n$$3 + \\frac{c}{4} = 3.025$$\n\nSubtracting $3$ from both sides gives:\n\n$$\\frac{c}{4} = 0.025 \\implies c = 0.1$$\n\nSince $c = 0.1$, the dopant concentration is positive ($c > 0$)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$N(c) = \\left(\\htmlData{anchor=first-term}{\\color{@sky}{3}} + \\htmlData{anchor=second-term}{\\color{@amber}{\\frac{c}{4}}}\\right)^9$"},{"text":"Every term in the binomial expansion of this expression is of the form:"},{"latex":"$$T_{k+1} = \\binom{9}{k} (\\color{@sky}{3})^{9-k} \\left(\\color{@amber}{\\frac{c}{4}}\\right)^k$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"first-term"},{"color":"amber","style":"text-color","anchor":"second-term"}],"connections":[]},"order":1,"title":"Analyze the expansion terms","description":"Since the constant term $3 > 0$ and $c = 0.1 > 0$, both parts of the binomial expression are positive. Because the binomial coefficients $\\binom{9}{k}$ are also positive, every single term in the full expansion must be positive."},{"type":"text","order":2,"title":"Determine the approximation type","content":"The approximation in part (b) only uses the first few terms of the expansion. The exact value of $3.025^9$ is equal to:\n\n$$\\text{Exact Value} = \\text{Approximation} + \\sum_{k=4}^{9} \\binom{9}{k} 3^{9-k} \\left(\\frac{c}{4}\\right)^k$$\n\nSince all the omitted higher-order terms ($c^4, c^5, \\dots, c^9$) are strictly positive, the sum of these omitted terms is positive. \n\nTherefore, the approximation is an **underestimate** because omitting these positive terms makes our approximated value smaller than the exact value."}]}],"generatedAt":"2026-06-09T04:18:09.081Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"3395870a-9db7-447e-915f-533c2459d67d","revisionId":"d397f2ec-553f-4533-bbc6-ea90b6399060","questionNumber":null,"questionType":"LA","specification":"In the expansion of $(1 + ax)^6$, where $a$ is a positive constant, the coefficient of $x^2$ is $60$.","difficulty":null,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"6f573365-5e0a-45a3-89b1-47e8ee0e91f5","label":"a","content":"Find the value of $a$.","markscheme":"- Coefficient of $x^2$ is $\\binom{6}{2}a^2 = 15a^2$ M1\n- $15a^2 = 60$, so $a^2 = 4$ M1\n- Since $a > 0$, $a = 2$ A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]},{"id":"f47f0701-bb97-4840-bcc9-a0a096223b41","label":"b","content":"Hence find the coefficient of $x^3$ in the expansion.","markscheme":"- Coefficient of $x^3$ is $\\binom{6}{3}a^3 = 20 \\times 8$ M1\n- $= 160$ A1\n\nFollow through from their value of $a$.","marks":2,"order":1,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"37dac930-d8b4-4b94-b373-099450f17b18","revisionId":"86e9801c-7d7a-4c83-8bf5-707875f257fb","questionNumber":null,"questionType":"LA","specification":"","difficulty":6,"difficultyBand":"medium","totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"dc9252ba-fb3a-4d73-9b61-dc3be8d082ae","label":"a","content":"Determine the first two terms, in ascending powers of \$u\$, of the binomial expansion of\n\\[\\left(1 + \\frac{3}{4}u\\right)^{-\\frac{1}{3}}\\]","markscheme":"- Applies binomial expansion $(1 + z)^n \\approx 1 + nz$ with $n = -\\frac{1}{3}$ and $z = \\frac{3}{4}u$ M1\n- Simplifies to obtain the expansion $1 - \\frac{1}{4}u$ A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"binomial_expansion","label":"Binomial Expansion Formula","steps":[{"type":"text","order":0,"title":"Identify the formula","content":"To find the first two terms of the binomial expansion, we use the binomial series expansion formula from

1.7 D: Sequences and series

Question 113

1.7 D: Sequences and series Questions

1.7 D: Sequences and series

Question 113

1.7 D: Sequences and series Questions