# The sum of the series $\sum_{n=1}^{\infty } \frac{n^2+6n +10}{(2n+1)!}$ is equal to : Option: 1 $\frac{41}{8}e+\frac{19}{8}e^{-1}+10$ Option: 2 $\frac{41}{8}e-\frac{19}{8}e^{-1}-10$ Option: 3 $\frac{41}{8}e+\frac{19}{8}e^{-1}-10$ Option: 4 $\frac{41}{8}e+\frac{19}{8}e^{-1}-10$

$T_{n}=\frac{n^{2}+6 n+10}{(2 n+1) !}=\frac{4 n^{2}+24 n+40}{4 \cdot(2 n+1) !}$

$\\=\frac{(2 n+1)^{2}+20 n+39}{4 \cdot(2 n+1) !} \\ =\frac{(2 n+1)^{2}+(2 n+1) \cdot 10+29}{4(2 n+1) !} \\ =\frac{1}{4}\left[\frac{(2 n+1)^{2}}{(2 n+1)(2 n) !}+\frac{(2 n+1) 10}{(2 n+1)(2 n) !}+\frac{29}{(2 n+1) !}\right] \\ =\frac{1}{4}\left[\frac{2 n+1}{(2 n) !}+\frac{10}{(2 n) !}+\frac{29}{(2 n+1) !}\right] \\ =\frac{1}{4}\left[\frac{1}{(2 n-1) !}+\frac{11}{(2 n) !}+\frac{29}{(2 n+1) !}\right]$

$\\\mathrm{S}_{1}=\sum\frac{1}{(2n-1)!}=\frac{1}{1 !}+\frac{1}{3 !}+\frac{1}{5 !}+\ldots=\frac{\mathrm{e}-\frac{1}{\mathrm{e}}}{2}\\\mathrm{S}_{2}=\sum\frac{11}{(2n)!}=11\left[\frac{1}{2 !}+\frac{1}{4 !}+\frac{1}{6 !}+\ldots\right]=11\left[\frac{\mathrm{e}+\frac{1}{\mathrm{e}}-2}{2}\right]\\\mathrm{S}_{3}=\sum\frac{29}{(2n-1)!}=29\left[\frac{1}{3 !}+\frac{1}{5 !}+\frac{1}{7 !}+\ldots\right]=29\left[\frac{\mathrm{e}-\frac{1}{\mathrm{e}}-2}{2}\right]$

$\\\text { Now, } \mathrm{S}=\frac{1}{4}\left[\mathrm{~S}_{1}+\mathrm{S}_{2}+\mathrm{S}_{3}\right] \\ =\frac{1}{4}\left[\frac{\mathrm{e}}{2}-\frac{1}{2 \mathrm{e}}+\frac{11 \mathrm{e}}{2}+\frac{11}{2 \mathrm{e}}+\frac{29 \mathrm{e}}{2}-\frac{29}{2 \mathrm{e}}-4\right] \\ =\frac{41 \mathrm{e}}{8}-\frac{19}{8 \mathrm{e}}-10$

Note That:

$\\e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\\\text{put x = 1}\\e=1+\frac{1}{1!}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots\\\text{put x}=-1\\e^{-1}=1-\frac{1}{1!}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}+\ldots$

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