A) \[{{e}^{2}}\]
B) \[e+{{e}^{-1}}\]
C) \[\frac{e-{{e}^{-1}}}{2}\]
D) \[\frac{3e-{{e}^{-1}}}{2}\]
Correct Answer: D
Solution :
Sum of series \[=1+2+\frac{1}{2!}+\frac{2}{3!}+\frac{1}{4!}+\frac{2}{5!}+.....\] \[=\left( 1+\frac{1}{2!}+\frac{1}{4!}+.... \right)+2\left( 1+\frac{1}{3!}+\frac{1}{5!}+..... \right)\] = \[\frac{e+{{e}^{-1}}}{2}+2.\frac{e-{{e}^{-1}}}{2}=\frac{3e-{{e}^{-1}}}{2}\].
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