A) \[e\]
B) \[{{e}^{-1}}\]
C) \[2e\]
D) \[{{e}^{2}}\]
Correct Answer: B
Solution :
Let\[{{I}_{n}}\]be the nth term of the given series. Then, \[{{T}_{n}}=\frac{2n}{(2n+1)!},n=1,2,3...\infty \] \[\therefore \] \[\left( \frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+...\infty \right)\] \[=\sum\limits_{n=1}^{\infty }{{{T}_{n}}}=\sum\limits_{n=1}^{\infty }{\frac{2n}{(2n+1)!}}=\sum\limits_{n=1}^{\infty }{\frac{(2n+1-1)}{(2n+1)!}}\] \[=\sum\limits_{n=1}^{\infty }{\frac{(2n+1)}{(2n+1)!}}-\sum\limits_{n=1}^{\infty }{\frac{1}{(2n+1)!}}\] \[=\sum\limits_{n=1}^{\infty }{\frac{1}{(2n)!}}-\sum\limits_{n=1}^{\infty }{\frac{1}{(2n+1)!}}\] \[=\left( \frac{e+{{e}^{-1}}}{2}-1 \right)-\left( \frac{e-{{e}^{-1}}}{2}-1 \right)={{e}^{-1}}\]
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