Sequence And Series Question 426
Question: $ 1+\frac{1+x}{2,!}+\frac{1+x+x^{2}}{3,!}+\frac{1+x+x^{2}+x^{3}}{4,!}+…..\infty = $
Options:
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} $
Show Answer
Answer:
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}{ \frac{1}{n!}x^{n}-\frac{1}{n!} } $ $ \sum\limits_{n=1}^{\infty }{T_{n}=\frac{1}{x-1}{ \sum\limits_{n=1}^{\infty }{,\frac{x^{n}}{n!}-}\sum\limits_{n=1}^{\infty }{,\frac{1}{n!}-} }}=\frac{1}{x-1}(e^{x}-e) $ .
BETA
