A) \[\frac{{{e}^{x}}+1}{x+1}\]
B) \[\frac{{{e}^{x}}+1}{x-1}\]
C) \[\frac{{{e}^{x}}-e}{x+1}\]
D) \[\frac{{{e}^{x}}-e}{x-1}\]
Correct Answer: D
Solution :
\[{{T}_{n}}=\frac{1+x+{{x}^{2}}+....+{{x}^{n-1}}}{n!}=\frac{1-{{x}^{n}}}{1-x}.\frac{1}{n!}\] \[=\frac{1}{x-1}\left\{ \frac{1}{n!}{{x}^{n}}-\frac{1}{n!} \right\}\] \[\sum\limits_{n=1}^{\infty }{{{T}_{n}}=\frac{1}{x-1}\left\{ \sum\limits_{n=1}^{\infty }{\,\frac{{{x}^{n}}}{n!}-}\sum\limits_{n=1}^{\infty }{\,\frac{1}{n!}-} \right\}}=\frac{1}{x-1}({{e}^{x}}-e)\].
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