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

  2. $\frac{41}{8} \mathrm{e}-\frac{19}{8} \mathrm{e}^{-1}-10$

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

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


Correct Option: , 2

Solution:

$\mathrm{T}_{\mathrm{n}}=\frac{\mathrm{n}^{2}+6 \mathrm{n}+10}{(2 \mathrm{n}+1) !}=\frac{4 \mathrm{n}^{2}+24 \mathrm{n}+40}{4(2 \mathrm{n}+1) !}$

$=\frac{(2 n+1)^{2}+20 n+39}{4 \cdot(2 n+1) !}$

$=\frac{(2 n+1)^{2}+(2 n+1) \cdot 10+29}{4(2 n+1) !}$.

$=\frac{1}{4}\left[\frac{(2 n+1)^{2}}{(2 n+1)(2 n) !}+\frac{(2 n+1) 10}{(2 n+1)(2 n) !}+\frac{29}{(2 n+1) !}\right]$

$=\frac{1}{4}\left[\frac{2 n+1}{(2 n) !}+\frac{10}{(2 n) !}+\frac{29}{(2 n+1) !}\right]$

$=\frac{1}{4}\left[\frac{1}{(2 n-1) !}+\frac{11}{(2 n) !}+\frac{29}{(2 n+1) !}\right]$

$\mathrm{S}_{1}=\frac{1}{1 !}+\frac{1}{3 !}+\frac{1}{5 !}+\ldots=\frac{\mathrm{e}-\frac{1}{\mathrm{e}}}{2}$

$\mathrm{S}_{2}=11\left[\frac{1}{2 !}+\frac{1}{4 !}+\frac{1}{6 !}+\ldots\right]=11\left[\frac{\mathrm{e}+\frac{1}{\mathrm{e}}-2}{2}\right]$

$S_{3}=29\left[\frac{1}{3 !}+\frac{1}{5 !}+\frac{1}{7 !}+\ldots\right]=29\left[\frac{\mathrm{e}-\frac{1}{\mathrm{e}}-2}{2}\right]$

Now, $S=\frac{1}{4}\left[S_{1}+S_{2}+S_{3}\right]$

$=\frac{1}{4}\left[\frac{\mathrm{e}}{2}-\frac{1}{2 \mathrm{e}}+\frac{11 \mathrm{e}}{2}+\frac{11}{2 \mathrm{e}}+\frac{29 \mathrm{e}}{2}-\frac{29}{2 \mathrm{e}}-4\right]$

$=\frac{41 \mathrm{e}}{8}-\frac{19}{8 \mathrm{e}}-10$

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