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 :
\[{{T}_{n}}=\frac{1+x+{{x}^{2}}+...+{{x}^{n-1}}}{n!}\] \[=\frac{{{x}^{n}}-1}{x-1}\cdot \frac{1}{n!}\] \[=\frac{1}{1-x}\left( \frac{{{x}^{n}}}{n!}-\frac{1}{n!} \right)\] \[\Rightarrow \] \[\sum\limits_{n=1}^{\infty }{\frac{1}{1-x}\left( \frac{{{x}^{n}}}{n!}-\frac{1}{n!} \right)}\] \[=\frac{1}{(x-1)}\left[ \sum\limits_{n=1}^{\infty }{\frac{{{x}^{n}}}{n!}-\sum\limits_{n=1}^{\infty }{\frac{1}{n!}}} \right]\] \[=\frac{1}{(x-1)}({{e}^{x}}-e)\]
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