A) \[\frac{{{2}^{n}}}{n!}\]
B) \[\frac{{{2}^{n}}-2}{n!}\]
C) \[\frac{{{2}^{n-1}}-1}{n!}\]
D) \[\frac{{{2}^{n-1}}}{n!}\]
Correct Answer: D
Solution :
\[{{\left( 1+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}+... \right)}^{2}}={{\left( \frac{{{e}^{x}}+{{e}^{-x}}}{2} \right)}^{2}}\] \[=\frac{1}{4}({{e}^{2x}}+{{e}^{-2x}}+2)\] \[\frac{1}{4}\left\{ 2\left( \frac{{{(2x)}^{2}}}{2!}+\frac{{{(2x)}^{4}}}{4!}+... \right)+2 \right\}\] So, coefficient of \[{{x}^{n}}\](n even) \[=\frac{1}{2}\left\{ \frac{{{2}^{n}}}{n!} \right\}=\frac{{{2}^{n-1}}}{n!}\]
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