A) \[1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+....\infty \]
B) 0
C) 1
D) none of the above
Correct Answer: A
Solution :
\[\frac{{{e}^{2}}+1}{2e}=\frac{e+{{e}^{-1}}}{2}\] \[=\frac{1}{2}\left[ \left( 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+... \right)+\left( 1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+... \right) \right]\] \[=\frac{1}{2}\left[ 2\left( 1+\frac{1}{2!}+\frac{1}{4!}+... \right) \right]\] \[=1+\frac{1}{2!}+\frac{1}{4!}+....\]
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