Sequence And Series Question 140
Question: The coefficient of $ x^{r} $ in the expansion of $ {e^{e^{x}}} $ is
Options:
A) $ \frac{1^{r}}{1,!}+\frac{2^{r}}{2,!}+\frac{3^{r}}{3,!}+….. $
B) $ 1+\frac{1}{1,!}+\frac{1}{2,!}+….+\frac{1}{r,!} $
C) $ \frac{1}{r,!}[ \frac{1^{r}}{1,!}+\frac{2^{r}}{2,!}+\frac{3^{r}}{3,!}+…. ] $
D) $ \frac{e^{r}}{r!} $
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Answer:
Correct Answer: C
Solution:
Expansion of $ {e^{e^{x}}} $ $ =1+\frac{e^{x}}{1!}+\frac{e^{2x}}{2!}+\frac{e^{3x}}{3!}+\frac{e^{4x}}{4!}+….+\frac{e^{rx}}{r!}+…. $ $ =1+\frac{1}{1!}[ 1+\frac{x}{1!}+\frac{x^{2}}{2!}+….+\frac{x^{r}}{r!}+….. ] $ $ +\frac{1}{2!}[ 1+\frac{2x}{1!}+\frac{{{(2x)}^{2}}}{2!}+….+\frac{{{(2x)}^{r}}}{r!}+….. ] $ + ??? Hence the coefficient of $ x^{r}=\frac{1}{r!}[ \frac{1^{r}}{1!}+\frac{2^{r}}{2!}+\frac{3^{r}}{3!}+\frac{4^{r}}{4!}+…. ] $ .
BETA
