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

We need to evaluate $$\displaystyle\sum_{n=1}^{\infty} \dfrac{n^2 + 6n + 10}{(2n+1)!}$$.

Substituting $$m = 2n + 1$$ so that $$n = \dfrac{m-1}{2}$$, we get $$$n^2 + 6n + 10 = \dfrac{(m-1)^2}{4} + 3(m-1) + 10 = \dfrac{m^2 + 10m + 29}{4}$$$. As $$n$$ runs from 1 to $$\infty$$, $$m$$ runs over odd values $$3, 5, 7, \ldots$$

So the sum becomes $$$S = \dfrac{1}{4}\displaystyle\sum_{\substack{m=3 \\ m \text{ odd}}}^{\infty} \dfrac{m^2 + 10m + 29}{m!}$$$.

We decompose $$$m^2 + 10m + 29 = m(m-1) + 11m + 29$$$ and use the standard sums over odd $$m \geq 3$$:

$$$\displaystyle\sum_{\substack{m=3\\m\text{ odd}}}^{\infty} \dfrac{1}{m!} = \dfrac{e - e^{-1}}{2} - 1$$$, since $$\sum_{\text{odd }m\geq 1}\dfrac{1}{m!} = \sinh(1) = \dfrac{e-e^{-1}}{2}$$ and we subtract the $$m=1$$ term.

$$$\displaystyle\sum_{\substack{m=3\\m\text{ odd}}}^{\infty} \dfrac{m}{m!} = \displaystyle\sum_{\substack{m=3\\m\text{ odd}}}^{\infty} \dfrac{1}{(m-1)!} = \displaystyle\sum_{k=1}^{\infty}\dfrac{1}{(2k)!} = \cosh(1) - 1 = \dfrac{e + e^{-1}}{2} - 1$$$.

$$$\displaystyle\sum_{\substack{m=3\\m\text{ odd}}}^{\infty} \dfrac{m(m-1)}{m!} = \displaystyle\sum_{\substack{m=3\\m\text{ odd}}}^{\infty} \dfrac{1}{(m-2)!} = \displaystyle\sum_{k=0}^{\infty}\dfrac{1}{(2k+1)!} = \sinh(1) = \dfrac{e - e^{-1}}{2}$$$.

Substituting: $$$S = \dfrac{1}{4}\left[\dfrac{e - e^{-1}}{2} + 11\left(\dfrac{e + e^{-1}}{2} - 1\right) + 29\left(\dfrac{e - e^{-1}}{2} - 1\right)\right]$$$.

Expanding: $$$S = \dfrac{1}{4}\left[\dfrac{e - e^{-1}}{2} + \dfrac{11e + 11e^{-1}}{2} - 11 + \dfrac{29e - 29e^{-1}}{2} - 29\right]$$$.

Combining the coefficients of $$e$$ and $$e^{-1}$$: the coefficient of $$e$$ is $$\dfrac{1 + 11 + 29}{2} = \dfrac{41}{2}$$, and the coefficient of $$e^{-1}$$ is $$\dfrac{-1 + 11 - 29}{2} = \dfrac{-19}{2}$$. The constant is $$-40$$.

Therefore $$$S = \dfrac{1}{4}\left[\dfrac{41e}{2} - \dfrac{19e^{-1}}{2} - 40\right] = \dfrac{41}{8}e - \dfrac{19}{8}e^{-1} - 10$$$.