• # question_answer If  $n$ is even, then in the expansion of  ${{\left( 1+\frac{{{x}^{2}}}{2\,!}+\frac{{{x}^{4}}}{4\,!}+...... \right)}^{2}}$, the coefficient of ${{x}^{n}}$ is A) $\frac{{{2}^{n}}}{n\,!}$ B) $\frac{{{2}^{n}}-2}{n\,\,!}$ C) $\frac{{{2}^{n-1}}-1}{n\,!}$ D) $\frac{{{2}^{n-1}}}{n\,!}$

${{\left( 1+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}+.... \right)}^{2}}={{\left( \frac{{{e}^{x}}+{{e}^{-x}}}{2} \right)}^{2}}$                                  $=\frac{1}{4}({{e}^{2x}}+{{e}^{-2x}}+2)$ $=\frac{1}{4}\left\{ 2\,\left( 1+\frac{{{(2x)}^{2}}}{2!}+\frac{{{(2x)}^{4}}}{4!}+.... \right)+2 \right\}$ \ The coefficient of  ${{x}^{n}}$(n even)$=\frac{1}{2}\left\{ \frac{{{2}^{n}}}{n!} \right\}=\frac{{{2}^{n-1}}}{n!}$.