JEE Main 2016 · 9 Apr · Shift 1 · Mathematics Q67

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MathematicsSingle correct+4 / -1Question 67 of 90

For $x\in\mathbb R$ , $x\ne-1$ , if $(1+x)^{2016}+x(1+x)^{2015}+x^2(1+x)^{2014}+\dots+x^{2016}=\displaystyle\sum_{i=0}^{2016}a_ix^i$ , then $a_{17}$ is equal to :

\dfrac{2017!}{17!\,2000!}

\dfrac{2016!}{17!\,1999!}

\dfrac{2017!}{2000!}

\dfrac{2016!}{16!}

Answer and worked solution

Official answer: A

$a_{17}=\sum_{k=0}^{17}\binom{2016-k}{17-k}=\binom{2017}{17}=\dfrac{2017!}{17!\,2000!}$ (hockey-stick identity).

Topics:Properties of binomial coefficients

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JEE Main 2016 · 9 Apr · Shift 1, Mathematics question 67