A) \[\frac{{{e}^{3}}-1}{2}\]
B) \[\frac{{{e}^{3}}-e}{2}\]
C) \[\frac{e-3}{2}\]
D) None of these
Correct Answer: B
Solution :
Given that \[{{T}_{n}}=\frac{1}{2}\left[ \frac{{{3}^{n}}}{n!}-\frac{1}{n!} \right]\] Therefore sum of the series \[\sum\limits_{n=1}^{\infty }{{{T}_{n}}=\frac{1}{2}\left[ \sum\limits_{n=1}^{\infty }{\,\frac{{{3}^{n}}}{n!}-\sum\limits_{n=1}^{\infty }{\,\frac{1}{n!}}} \right]}\] \[=\frac{1}{2}\left[ \left\{ 1+\frac{3}{1!}+\frac{{{3}^{2}}}{2!}+.... \right\}-\left\{ 1+\frac{1}{1!}+\frac{1}{2!}+.... \right\} \right]\]\[=\frac{1}{2}({{e}^{3}}-e)\].
You need to login to perform this action.
You will be redirected in
3 sec
