Binomial Theorem
Binomial Coefficients
Grade 11
Question:
<p>For \(x\in\mathbb{R}\), \(x\neq -1\), if \((1+x)^{2016}+x(1+x)^{2015}+x^2(1+x)^{2014}+\cdots+x^{2016}=\displaystyle\sum_{i=0}^{2016}a_ix^i\), then \(a_{17}\) is equal to</p>
<p>\(\dfrac{2017!}{17!\,2000!}\)</p>
<p>\(\dfrac{2016!}{17!\,1999!}\)</p>
<p>\(\dfrac{2016!}{16!}\)</p>
<p>\(\dfrac{2017!}{2000!}\)</p>
Step-by-Step Solution
Key Concept: Recognize the left side as a geometric series with first term (1+x)^2016 and common ratio x/(1+x), then use the closed form to extract the coefficient of x^17.
<p><strong>Step 1: Recognize the geometric series structure</strong></p><p>The sum S = (1+x)^2016 + x(1+x)^2015 + x²(1+x)^2014 + ... + x^2016 is a geometric series with:</p><p>• First term a = (1+x)^2016</p><p>• Common ratio r = x/(1+x)</p><p>• Number of terms = 2017</p><p><strong>Step 2: Apply geometric series formula</strong></p><p>S = (1+x)^2016 · [1 - (x/(1+x))^2017] / [1 - x/(1+x)]</p><p>S = (1+x)^2016 · [1 - x^2017/(1+x)^2017] / [(1+x-x)/(1+x)]</p><p>S = (1+x)^2016 · [(1+x)^2017 - x^2017] / (1+x)^2017</p><p>S = [(1+x)^2016((1+x)^2017 - x^2017)] / (1+x)^2017</p><p>S = (1+x)^2016 - x^2017/(1+x)</p><p><strong>Step 3: Find coefficient of x^17</strong></p><p>The term x^2017/(1+x) contributes only to powers ≥ 2017 in the expansion, so it doesn't affect x^17.</p><p>Therefore, a₁₇ comes entirely from the coefficient of x^17 in (1+x)^2016.</p><p>By the binomial theorem: (1+x)^2016 = Σ C(2016,k)x^k</p><p>∴ a₁₇ = C(2016, 17) = 2016!/(17! · 1999!)</p>
Correct Answer: A