Question

For $x\in R,x\ne -1$${(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. $\frac{2017!}{17!2000!}$
• B. $\frac{2016!}{17!1999!}$
• C. $\frac{2017!}{2000!}$
• D. $\frac{2016!}{16!}$

Answer 1

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

Given:

$\sum _{i=0}^{2016}{a}_i{x}^i={(1+x)}^{2016}+x{(1+x)}^{2015}+x^2{(1+x)}^{2014}+................+x^{2016}$

Using formula,

$\sum _{i=0}^{}{a}_i{x}^i={(1+x)}^n+x{(1+x)}^{n-1}+x^2{(1+x)}^{n-2}+x^3{(1+x)}^{n-3}+............+x^n$

$\sum _{i=0}^{}{a}_i{x}^i=$ $\frac{{(1+x)}^n(1-{\left(\frac{x}{1+x}\right)}^{n+1})}{1-\frac{x}{1+x}}$ $$

$\sum _{i=0}^{2016}{a}_i{x}^i=$$\frac{{(1+x)}^{2016}(1-{\left(\frac{x}{1+x}\right)}^{2017})}{1-\frac{x}{1+x}}$

$=\frac{\frac{{(1+x)}^{2016}}{1}-\frac{x^{2017}}{(1+x)}}{\frac{1+x-x}{1+x}}$

$=\frac{{(1+x)}^{2017}-x^{2017}}{1}$

$={(1+x)}^{2017}-x^{2017}$

Then, Value of $a_{17}=?$

We know that,

$a_r{=}^n{C}_r=\frac{n!}{(n-r)!r!}$

Then

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

$a_{17}{=}^{2017}{C}_{17}=\frac{2017!}{(2017-17)!17!}=\frac{2017!}{2000!\times 17!}$

Hence, this is the required value.