A) \[e(e+1)\]
B) \[e(1-e)\]
C) \[3e-1\]
D) \[3e\]
E) \[e(e-1)\]
Correct Answer: E
Solution :
\[S=1+3+7+15+...+{{T}_{n}}+...\infty \] \[S=1+3+7+...+{{T}_{n-1}}+{{T}_{n}}+...\infty \] On subtraction, we get \[{{T}_{n}}=1+2+4+8+....\] \[=\frac{1({{2}^{n}}-1)}{2-1}={{2}^{n}}-1\] \[\therefore \] \[T_{n}^{}=\frac{{{2}^{n}}-1}{n!}\] \[\sum\limits_{n=1}^{\infty }{{{T}_{n}}}=\sum\limits_{n=1}^{\infty }{\left( \frac{{{2}^{n}}}{n!}-\frac{1}{n!} \right)}\] \[={{e}^{2}}-e=e(e-1)\]
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