Question

For $x\in R,x\ne -1$, if $(1+x{)}^{2016}+x(1+x{)}^{2015}+x^2(1+x{)}^{2014}+...+x^{2016}=\sum _{i=0}^{2016}{a}_i{x}^i$, then $a_{17}$ is equal to:

• A. $\frac{2016!}{16!}$
• B. $\frac{2016!}{2000!}$
• C. $\frac{2016!}{17!1999!}$
• D. $\frac{2017!}{17!2000!}$

Answer 1

The correct option is D $\frac{2017!}{17!2000!}$

It is a GP with $(1+x{)}^{2016}$ as 1st term and $\frac{x}{1+x}$ as common ratio.

Total number of terms are $2017$

Sum $=\frac{a(r^n-1)}{r-1}$

$=\frac{(1+x{)}^{2016}[{\left(\frac{x}{1+x}\right)}^{2017}-1]}{(\frac{x}{1+x}-1)}$
$=-\left[\right(x{)}^{2017}-(1+x{)}^{2017}]$
$=(1+x{)}^{2017}-x^{2017}$

$a_{17}$ is basically the coefficient of $x^{17}$.

So, $a_{17}=$ ${}^{2017}{C}_{17}$