Question

The coefficient of $x^{r}$ in the expansion of $e^{e^{x}}$ is.

• 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. $\frac{1}{r!}\left\lbrack \frac{1^{r}}{1!} + \frac{2^{r}}{2!} + \frac{3^{r}}{3!} + .... \right\rbrack$
• D. $\frac{e^{r}}{r!}$

Answer 1

Answer: (C)

$\frac{1}{r!}\left\lbrack \frac{1^{r}}{1!} + \frac{2^{r}}{2!} + \frac{3^{r}}{3!} + .... \right\rbrack$

Answer 2

Expansion of $\frac{x}{1 - x} + \log_{e}(1 - x)$

$\frac{x - 1}{(x + 1)} + \frac{1}{2}.\frac{x^{2} - 1}{(x + 1)^{2}} + \frac{1}{3}.\frac{x^{3} - 1}{(x + 1)^{3}} + ......\infty =$

$\log_{e}x$

$\log_{e}(1 + x)$ + ………

Hence the coefficient of

$\log_{e}(1 - x)$.