A) \[\frac{{{e}^{-1}}}{2}\]
B) \[e\]
C) \[\frac{e}{4}\]
D) \[\frac{e}{6}\]
Correct Answer: A
Solution :
Let \[S=\frac{1}{3!}+\frac{2}{5!}+\frac{3}{7!}+.....\] \[\therefore \] \[{{T}_{n}}=\frac{n}{(2n+1)!}\] \[=\frac{1}{2}\left[ \frac{2n+1-1}{(2n+1)!} \right]=\frac{1}{2}\left[ \frac{1}{(2n)!}-\frac{1}{(2n+1)!} \right]\] \[\therefore \] \[{{T}_{1}}=\frac{1}{2}\left( \frac{1}{2!}-\frac{1}{3!} \right)\] \[{{T}_{2}}=\frac{1}{2}\left( \frac{1}{4!}-\frac{1}{5!} \right)\] ? ? ? ? ? ? \[\therefore \] \[S={{T}_{1}}+{{T}_{2}}+....\] \[=\frac{1}{2}\left[ \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+.....\infty +1-1 \right]\] \[=\frac{{{e}^{-1}}}{2}\]
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