If \[A=\sum\limits_{n=1}^{\infty }{\frac{2n}{(2n-1)!}B=\sum\limits_{n=1}^{\infty }{\frac{2n}{(2n+1)!}}}\] then AB is equal to

If \[A=\sum\limits_{n=1}^{\infty }{\frac{2n}{(2n-1)!}B=\sum\limits_{n=

1.	If \[A=\sum\limits_{n=1}^{\infty }{\frac{2n}{(2n-1)!}B=\sum\limits_{n=1}^{\infty }{\frac{2n}{(2n+1)!}}}\] then AB is equal to
A.	\[{{e}^{2}}\]
B.	e
C.	\[e+{{e}^{2}}\]
D.	1
Answer» E.

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If \[A=\sum\limits_{n=1}^{\infty }{\frac{2n}{(2n-1)!}B=\sum\limits_{n=1}^{\infty }{\frac{2n}{(2n+1)!}}}\] then AB is equal to

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