JEE Main 2016 (Online) 9th AprilQuestion 1
For $x\in R$, $x\neq-1$, if $(1+x)^{2016}+x(1+x)^{2015}+x^{2}(1+x)^{2014}+\ldots+x^{2016}=\sum_{i=0}^{2016}a_{i}x^{i}$, then $a_{17}$ is equal to
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. With $a=1+x$, $b=x$ we have $a-b=1$, so the sum closes to $a^{2017}-b^{2017}$. Formulas used. (1) $a^n+a^{n-1}b+\cdots+b^n=\dfrac{a^{n+1}-b^{n+1}}{a-b}$Step 1 - Close the sum. By (1) with $n=2016$: $$\text{sum}=(1+x)^{2017}-x^{2017}.$$ Step 2 - $a_{17}$. $x^{2017}$ does not affect $x^{17}$, so $$a_{17}={}^{2017}C_{17}=\frac{2017!}{17!\,2000!}.$$ Answer: (A)