# The sum of the series

Question:

The sum of the series $\sum_{n=1}^{\infty} \frac{n^{2}+6 n+10}{(2 n+1) !}$ is equal to:

1. (1) $\frac{41}{8} e+\frac{19}{8} e^{-1}-10$

2. (2) $-\frac{41}{8} e+\frac{19}{8} e^{-1}-10$

3. (3) $\frac{41}{8} e-\frac{19}{8} e^{-1}-10$

4. (4) $\frac{41}{8} e+\frac{19}{8} e^{-1}+10$

Correct Option: , 3

Solution:

$\sum_{n=1}^{\infty} \frac{n^{2}+6 n+10}{(2 n+1) !}$

Put $2 n+1=r$, where $r=3,5,7, \ldots$

$\Rightarrow n=\frac{r-1}{2}$

$\frac{n^{2}-6 n+10}{(2 n+1) !}=\frac{\left(\frac{r-1}{2}\right)^{2}+3 r-3+10}{r !}=\frac{r^{2}+10 r+29}{4 r !}$

Now $\sum_{r-3,5,7} \frac{r(r-1)+11 r+29}{4 r !}=\frac{1}{4} \sum_{r=3,5,7 \ldots \ldots . .}\left(\frac{1}{(r-2) !}+\frac{11}{(r-1) !}+\frac{29}{r !}\right)$

$=\frac{1}{4}\left\{\left(\frac{1}{1 !}+\frac{1}{3 !}+\frac{1}{5 !}+\ldots . . \cdot\right)+11\left(\frac{1}{2 !}+\frac{1}{4 !}+\frac{1}{6 !}+\ldots\right)+29\left(\frac{1}{3 !}+\frac{1}{5 !}+\frac{1}{7 !}+\ldots \ldots\right)\right\}$

$=\frac{1}{4}\left\{\frac{e-\frac{1}{2}}{2}+11\left(\frac{c+\frac{1}{2}-2}{2}\right)+29\left(\frac{\epsilon-\frac{1}{2}-2}{2}\right)\right\}$

$=\frac{1}{8}\left\{e-\frac{1}{e}+11 e+\frac{11}{e}-22+29 e-\frac{29}{e}-58\right\}$

$=\frac{1}{8}\left\{41 e-\frac{19}{6}-80\right\}$