A) \[{{e}^{2}}\]
B) e
C) \[e+{{e}^{2}}\]
D) 1
Correct Answer: D
Solution :
[d] \[A=\sum\limits_{n=1}^{\infty }{\frac{2n-1+1}{(2n-1)!}}\] \[=\sum\limits_{n=1}^{\infty }{\left[ \frac{1}{(2n-2)!}+\frac{1}{(2n-1)!} \right]}\] \[=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+....=e\] Similarly \[B={{e}^{-1}}\] as terms will be alternately positive and negative. \[\therefore \,\,\,\,\,AB=e\,.\,{{e}^{-1}}=1\]
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