• # question_answer In  the expansion of  $\frac{1+x}{1\,!}+\frac{{{(1+x)}^{2}}}{2\,!}+\frac{{{(1+x)}^{3}}}{3\,!}+.....,$ the coefficient of ${{x}^{n}}$ will be A) $\frac{1}{n\,!}$ B) $\frac{1}{n\,!}+\frac{1}{(n+1)\,!}$ C) $\frac{e}{n\,!}$ D) $e\,\left[ \frac{1}{n\,!}+\frac{1}{(n+1)\,!} \right]$

$\frac{1+x}{1!}+\frac{{{(1+x)}^{2}}}{2!}+\frac{{{(1+x)}^{3}}}{3!}+....\infty$ $={{e}^{1+x}}-1=e.\,\,{{e}^{x}}-1$ $=-1+e\left\{ 1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+.... \right\}$ \ The coefficient of  ${{x}^{n}}=e\frac{1}{n!}$.