Sequence And Series Question 538
Question: In the expansion of $ \frac{1+x}{1,!}+\frac{{{(1+x)}^{2}}}{2,!}+\frac{{{(1+x)}^{3}}}{3,!}+….., $ the coefficient of $ x^{n} $ will be
Options:
A) $ \frac{1}{n,!} $
B) $ \frac{1}{n,!}+\frac{1}{(n+1),!} $
C) $ \frac{e}{n,!} $
D) $ e,[ \frac{1}{n,!}+\frac{1}{(n+1),!} ] $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{1+x}{1!}+\frac{{{(1+x)}^{2}}}{2!}+\frac{{{(1+x)}^{3}}}{3!}+….\infty $ $ ={e^{1+x}}-1=e.e^{x}-1 $ $ =-1+e{ 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…. } $ \ The coefficient of $ x^{n}=e\frac{1}{n!} $ .
BETA
