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

we have , 

$(1+x{)}^{2016}+x(1+x{)}^{2015}+x^2(1+x{)}^{2014}+\dots +x^{2016}=\sum _{i=0}^{2016} a_i{x}^i$ 

$\Rightarrow (1+x{)}^{2016}\left(\frac{{\left({\displaystyle \frac{x}{1+x}}\right)}^{2017}-1}{\frac{x}{1+x}-1}\right)=\sum _{i=1}^{2016} a_i{x}^i$

$\Rightarrow \left\{(1+x{)}^{2017}-x^{2017}\right\}=\sum _{i=1}^{2016} a_i{x}^i$ 

$\Rightarrow a_{17}=$Coefficient of $x^{17}$ in $\left\{(1+x{)}^{2017}-x^{2017}\right\}$

$\Rightarrow a_{17}=$Coefficient of $x^{17}$ in  $(1+x{)}^{2017}$

$\Rightarrow a_{17}=$${ }^{2017}{C}_{17}=\frac{2017!}{17!2000!}$