Question

The sum of the series $\sum _{\mathrm{n}=1}^{\infty }\frac{\left[{\mathrm{n}}^2+6\mathrm{n}+10\right]}{(2\mathrm{n}+1)!}$ is equal to:

• A. $\left(\frac{41}{8}\right)\mathrm{e}+\left(\frac{19}{8}\right){\mathrm{e}}^{-1}-10$
• B. $\left(-\frac{41}{8}\right)\mathrm{e}+\left(\frac{19}{8}\right){\mathrm{e}}^{-1}-10$
• C. $\left(\frac{41}{8}\right)\mathrm{e}-\left(\frac{19}{8}\right){\mathrm{e}}^{-1}-10$
• D. $\left(\frac{41}{8}\right)\mathrm{e}+\left(\frac{19}{8}\right){\mathrm{e}}^{-1}+10$

Answer 1

The correct option is C

$\left(\frac{41}{8}\right)\mathrm{e}-\left(\frac{19}{8}\right){\mathrm{e}}^{-1}-10$

Explanation of the correct option:

We know that $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$ and $e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...$

Given : $\sum _{\mathrm{n}=1}^{\infty }\frac{\left[{\mathrm{n}}^2+6\mathrm{n}+10\right]}{(2\mathrm{n}+1)!}$

Put $2\mathrm{n}+1=\mathrm{r}$, where $\mathrm{r}=3,5,7......$

Thus $\begin{array}{rcl}\frac{\left[{\mathrm{n}}^2+6\mathrm{n}+10\right]}{(2\mathrm{n}+1)!}& =& \frac{{\left({\displaystyle \frac{r-1}{2}}\right)}^2+3r-3+10}{r!}\\ & =& \frac{r^2+10r+29}{4(r!)}\end{array}$

Now,

,$$\begin{gathered} \begin{array}{rcl}\sum _{\mathrm{n}=1}^{\infty }\frac{\left[{\mathrm{n}}^2+6\mathrm{n}+10\right]}{(2\mathrm{n}+1)!}& =& \sum _{\mathrm{r}=3,5,7...}\frac{\mathrm{r}(\mathrm{r}-1)+11\mathrm{r}+29}{4(\mathrm{r}!)}\\ & =& \frac{1}{4}\sum _{\mathrm{r}=3,5,7...}\left(\frac{1}{(\mathrm{r}-2)!}+\frac{11}{(\mathrm{r}-1)!}+\frac{29}{\mathrm{r}!}\right)\\ & =& \frac{1}{4}\left\{\left(\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+......\right)+11\left(\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+.....\right)+29\left(\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+......\right)\right\}\\ & & \\ & =& \frac{1}{4}\left\{\frac{\mathrm{e}-{\displaystyle \frac{{\displaystyle 1}}{{\displaystyle \mathrm{e}}}}}{2}+11\left(\frac{\mathrm{e}+{\displaystyle \frac{1}{\mathrm{e}}}-2}{2}\right)+29\left(\frac{\mathrm{e}-{\displaystyle \frac{1}{\mathrm{e}}}-2}{2}\right)\right\}\left[\because e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...,e^{-1}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...\right]\\ & =& \frac{1}{8}\left\{\mathrm{e}-\frac{1}{\mathrm{e}}+11\mathrm{e}+\frac{11}{\mathrm{e}}-22+29\mathrm{e}-\frac{29}{\mathrm{e}}-58 \\ \right\}\\ & =& \frac{1}{8}\left\{41\mathrm{e}-\frac{19}{\mathrm{e}}-80\right\}\\ & =& \frac{41}{8}\mathrm{e}-\frac{19}{8}{\mathrm{e}}^{-1}-10\end{array} \end{gathered}$$Therefore, the value of $\sum _{\mathrm{n}=1}^{\infty }\frac{\left[{\mathrm{n}}^2+6\mathrm{n}+10\right]}{(2\mathrm{n}+1)!}$ is $\begin{array}{rcl}& & \frac{41}{8}\mathrm{e}-\frac{19}{8}{\mathrm{e}}^{-1}-10\end{array}$.

Hence, option (C) is the correct option.