A) \[{{e}^{2}}\]
B) \[\frac{1}{2}{{(e+{{e}^{-1}})}^{2}}\]
C) \[\frac{3e-1}{6}\]
D) \[\frac{4e+1}{6}\]
Correct Answer: C
Solution :
\[{{T}_{n}}=\frac{n(n+1)(2n+1)}{6n(n+1)!}\] \[\therefore S=\frac{1}{6}\sum{\left[ \frac{2n+1}{(n)!} \right]}\] = \[\frac{1}{6}\sum{\left[ 2.\frac{1}{(n-1)!}+\frac{1}{(n)!} \right]}\] = \[\frac{1}{6}[2.e+e-1]=\frac{1}{6}[3e-1]\].You need to login to perform this action.
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