A) \[1+\frac{2}{2\,!}+\frac{{{2}^{2}}}{3\,!}+\frac{{{2}^{3}}}{4\,!}+.....\infty \]
B) \[1+\frac{1}{2\,!}+\frac{1}{4\,!}+\frac{1}{6\,!}+.....\infty \]
C) \[\frac{1}{2}\left( 1+\frac{1}{2\,!}+\frac{1}{4\,!}+....\infty \right)\]
D) \[\frac{1}{2}\left( 1+\frac{1}{1\,!}+\frac{1}{2\,!}+\frac{1}{3\,!}+....\infty \right)\]
Correct Answer: B
Solution :
\[\frac{{{e}^{2}}+1}{2e}=\frac{e+{{e}^{-1}}}{2}=1+\frac{1}{2\ !\,}+\frac{1}{4\ !}+\frac{1}{6\ !}+......\infty \]
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