A) e
B) \[{{e}^{-\,\frac{1}{2}}}\]
C) \[{{e}^{-\,2}}\]
D) None of these
Correct Answer: D
Solution :
We know, \[{{e}^{x}}=1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+\frac{{{x}^{4}}}{4!}+....\] Put \[x=-1\]\[\Rightarrow {{e}^{-1}}=1-1+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+.....\] Þ \[{{e}^{-1}}=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+.....\]
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