A) \[\frac{{{e}^{x}}}{n!}\]
B) \[\frac{{{n}^{n}}}{n!}\]
C) \[\frac{1}{n!}\]
D) None of these
Correct Answer: D
Solution :
[d] Let \[{{e}^{x}}=z\], then |
\[{{e}^{{{e}^{x}}}}={{e}^{z}}=\sum\limits_{k=0}^{\infty }{\frac{{{z}^{k}}}{k!}=\sum\limits_{k=0}^{\infty }{\frac{{{({{e}^{x}})}^{k}}}{k!}=\sum\limits_{k=0}^{\infty }{\frac{{{e}^{kx}}}{k!}}}}\] |
\[=\left( 1+\frac{{{e}^{x}}}{1!}+\frac{{{e}^{2x}}}{2!}+\frac{{{e}^{3x}}}{3!}+....to\,\,\infty \right)\] |
\[=1+\frac{1}{1!}\left( \sum\limits_{n=0}^{\infty }{\frac{{{x}^{n}}}{n!}} \right)+\frac{1}{2!}\left( \sum\limits_{n=0}^{\infty }{\frac{{{(2x)}^{n}}}{n!}} \right)+\] |
\[\frac{1}{3!}\left( \sum\limits_{n=0}^{\infty }{\frac{{{(3x)}^{n}}}{n!}} \right)+to....\infty \] |
\[\therefore \] Coefficient of \[{{x}^{n}}\] in \[{{e}^{{{e}^{x}}}}\] |
\[=\frac{1}{1!}\left( \frac{1}{n!} \right)+\frac{1}{2!}\left( \frac{{{2}^{n}}}{n!} \right)+\frac{1}{3!}\left( \frac{{{3}^{n}}}{n!} \right)+....to\,\,\infty \] |
\[=\frac{1}{n!}\left( \frac{1}{1!}+\frac{{{2}^{n}}}{2!}+\frac{{{3}^{n}}}{3!}+.....to\,\,\infty \right)\] |
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