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

$T_n=\frac{n^2+6n+10}{\left(2n+1\right)!}=\frac{4n^2+24n+40}{4.\left(2n+1\right)!}$

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

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

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

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

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

$S_1=\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+\dots =\frac{e-\frac{1}{e}}{2}$

$S_2=11\left[\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\dots \right]=11\left[\frac{\left.e+\frac{1}{e}-2\right]}{2}\right]$

$S_3=29\left[\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+\dots \right]=29\left[\frac{e-\frac{1}{e}-2}{2}\right]$

Now, $S=\frac{1}{4}\left[S_1+S_2+S_3\right]$

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

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