Summation Challenge: Evaluate $\sum_{n=0}^\infty \frac{16n^2+20n+7}{(4n+2)!}$

[anemone]

Evaluate $\displaystyle \sum_{n=0}^\infty \dfrac{16n^2+20n+7}{(4n+2)!}$.

 

[anemone]

Spoiler: Solution of other

Consider the following two power series,

$\displaystyle \sin x=\sum_{n=0}^\infty (-1)^n\cdot \dfrac{x^{2n+1}}{(2n+1)!}$ and $\displaystyle e^x=\sum_{n=0}^\infty \dfrac{x^{n}}{n!}$, $x\in \mathbb{R} $

Hence we have

$\displaystyle \sin 1=\sum_{n=0}^\infty (-1)^n\cdot \dfrac{1}{(2n+1)!}=\sum_{n=0}^\infty \left(\dfrac{1}{(4n+1)!}-\dfrac{1}{(4n+3)!}\right)$ and

$\displaystyle e=\sum_{n=0}^\infty \dfrac{1}{n!}$

Using the above considerations we get

$\displaystyle \begin{align*} \sum_{n=0}^\infty \dfrac{16n^2+20n+7}{(4n+2)!}&=\sum_{n=0}^\infty \dfrac{(4n+2)(4n+1)+2(4n+2)+1}{(4n+2)!}\\&=\sum_{n=0}^\infty \left(\dfrac{1}{(4n)!}+\dfrac{2}{(4n+1)!}+\dfrac{1}{(4n+2)!}\right)\\&=\sum_{n=0}^\infty \dfrac{1}{n!}+\sum_{n=0}^\infty \left(\dfrac{1}{(4n+1)!}-\dfrac{1}{(4n+3)!}\right)\\&=e+\sin 1\end{align*}$