Binomial Theorem And Its Simple Applications Question 53
Question: The coefficient of $ x^{n} $ in the expansion of $ {e^{e^{x}}} $ is
Options:
A) $ \frac{e^{x}}{n!} $
B) $ \frac{n^{n}}{n!} $
C) $ \frac{1}{n!} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] Let $ e^{x}=z $ , then $ {e^{e^{x}}}=e^{z}=\sum\limits_{k=0}^{\infty }{\frac{z^{k}}{k!}=\sum\limits_{k=0}^{\infty }{\frac{{{(e^{x})}^{k}}}{k!}=\sum\limits_{k=0}^{\infty }{\frac{e^{kx}}{k!}}}} $
$ =( 1+\frac{e^{x}}{1!}+\frac{e^{2x}}{2!}+\frac{e^{3x}}{3!}+….to\infty ) $
$ =1+\frac{1}{1!}( \sum\limits_{n=0}^{\infty }{\frac{x^{n}}{n!}} )+\frac{1}{2!}( \sum\limits_{n=0}^{\infty }{\frac{{{(2x)}^{n}}}{n!}} )+ $
$ \frac{1}{3!}( \sum\limits_{n=0}^{\infty }{\frac{{{(3x)}^{n}}}{n!}} )+to….\infty $
$ \therefore $ Coefficient of $ x^{n} $ in $ {e^{e^{x}}} $
$ =\frac{1}{1!}( \frac{1}{n!} )+\frac{1}{2!}( \frac{2^{n}}{n!} )+\frac{1}{3!}( \frac{3^{n}}{n!} )+….to\infty $
$ =\frac{1}{n!}( \frac{1}{1!}+\frac{2^{n}}{2!}+\frac{3^{n}}{3!}+…..to\infty ) $
BETA
