Find the value of the expression involving $e$ 
Find the value of the expression $$\sum_{n=1}^{\infty}\frac{n^2}{(2n-2)!}$$

I was doing a problem in which I got this expression as a sub problem.
I donated my brain for over $2$ hours to this question but in vain. I also saw some calculators, but they are solving it with the help of hyperbolic trigonometry which I don't know yet.
I only know expansion of $e$. The original question was

$$\sum_{n=0}^{\infty}\frac{n^3((2n)!+(2n-1)(n!))}{((2n)!)(n!)}$$

Any help is greatly appreciated.
 A: Little introduction to hyperbolic geometry :
Let $\operatorname{ch}(x)=\frac{e^x+e^{-x}}{2}$ and $\operatorname{sh}(x)=\frac{e^x-e^{-x}}{2}$. Using the Taylor expansion of $\exp$, namely $e^x=\sum_{n=0}^{+\infty}\frac{x^n}{n!}$ we get that
$$ \operatorname{ch}(x)=\sum_{n=0}^{+\infty}\frac{x^{2n}}{(2n)!} \text{ and } \operatorname{sh}(x)=\sum_{n=0}^{+\infty}\frac{x^{2n+1}}{(2n+1)!} $$
Back to the question :
On the one hand,
$$ \sum_{n=1}^{+\infty}\frac{2n(2n-1)}{(2n)!}=\sum_{n=1}^{+\infty}\frac{1}{(2n-2)!}=\operatorname{ch}(1) $$
on the other hand,
$$ \sum_{n=1}^{+\infty}\frac{2n(2n-1)}{(2n)!}=4\sum_{n=1}^{+\infty}\frac{n^2}{(2n)!}-\sum_{n=1}^{+\infty}\frac{1}{(2n-1)!}=4\sum_{n=1}^{+\infty}\frac{n^2}{(2n)!}-\operatorname{sh}(1) $$
thus
$$ \sum_{n=1}^{+\infty}\frac{n^2}{(2n)!}=\frac{\operatorname{ch}(1)+\operatorname{sh}(1)}{4}=\frac{e}{4}. $$
This means that
$$ \sum_{n=0}^{+\infty}\frac{(n+1)^2}{(2n)!}=\sum_{n=1}^{+\infty}\frac{n^2}{(2n)!}+\operatorname{sh}(1)+\operatorname{ch}(1)=\frac{5e}{4}. $$
Remark :
I've used hyperbolic trigonometry only to lighten the calculations, but you can of course replace $\operatorname{ch}$ and $\operatorname{sh}$ with their expressions using exponentials.
A: $4n^2=(2n-2)(2n-3)+5(2n-2)+4$ and for $n=1,$ this reduces to $4=0+5\cdot0+4,$ hence
$$4\sum_{n=1}^{\infty}\frac{n^2}{(2n-2)!}=\sum_{n=2}^\infty\frac1{(2n-4)!}+ \sum_{n=2}^\infty\frac5{(2n-3)!}+\sum_{n=1}^\infty\frac4{(2n-2)!}$$
$$=\sum_{k=0}^\infty\frac{1+4}{(2k)!}+\sum_{k=0}^\infty\frac5{(2k+1)!}$$
$$=5\sum_{n=0}^\infty\frac1{n!}=5e.$$
Therefore,
$$\sum_{n=1}^{\infty}\frac{n^2}{(2n-2)!}=\frac{5e}4.$$
A: Let
$$f(x)=\sum_{n=1}^\infty \frac{x^n}{(2n-2)!}=\frac{x(e^{\sqrt x}+ e^{-\sqrt x })}2$$
Then
$$x f'(x)=x \sum_{n=1}^\infty \frac{n x^{n-1}}{(2n-2)!}=x \sum_{n=1}^\infty \frac{n x^n}{(2n-2)!}$$
and
$$(x f'(x))'=\sum_{n=1}^\infty \frac{n^2 x^{n-1}}{(2n-2)!}$$
Now plug in $x=1$ into $(x f'(x))'=f'(x)+xf''(x)$. This gives us the answer $\frac{5e}4$.
