<s>[INST] How to evaluate $1+\frac{2^2}{3!}+\frac{3^2}{5!}+\frac{4^2}{7!}+\cdots$? I learnt that $\displaystyle \sum_{n=0}^{\infty} \frac{n+1}{(2n+1)!} = \frac{e}{2}$.
I am wondering what the closed form for $\displaystyle \sum_{n=0}^{\infty} \frac{(n+1)^2}{(2n+1)!}$ is. 
I tried using the fact that $ 1+3+5+\cdots+(2n-1) = n^2$, but it was not fruitful.
Could people give me some hints on how to approach this problem?
To view the general formula, please visit The value of $\sum_{n=0}^{\infty}\frac{(n+1)^k}{(2n+1)!}$, where $k\in\mathbb{W}$
 [/INST] 
$$\sum_{n=0, \text{odd}}^{\infty} \frac{(\frac{n+1}{2})^2}{n!}$$

$$=\frac{1}{4} \sum_{n=0, \text{odd}}^{\infty} \frac{n^2+2n+1}{n!}$$
$$=\frac{1}{4} \left( \sum_{n=0,\text{odd}}^{\infty} \frac{n}{(n-1)!}+2\sum_{n=0,\text{odd}}^{\infty} \frac{1}{(n-1)!}+\sum_{n=0,\text{odd}}^{\infty} \frac{1}{n!} \right)$$

Now note:
$$\sum_{n=0, \text{odd}}^{\infty} \frac{1}{n!}$$
$$=\sum_{n=0}^{\infty} \frac{1}{2} \frac{1^n-(-1)^n}{n!} $$
$$=\frac{e^{1}-e^{-1}}{2}= \sinh 1$$

Also note:
$$S :=\sum_{n=0, \text{odd}}^{\infty} \frac{1}{(n-1)!}$$
$$=\sum_{n=0,\text{even}}^{\infty} \frac{1}{n!}$$
$$=\sum_{n=0}^{\infty}  \frac{1}{2} \frac{1+(-1)^n}{n!}$$
$$=\cosh 1$$

Finally,
$$\sum_{n=0, \text{odd}}^{\infty} \frac{n}{(n-1)!}$$
$$=\sum_{n=0,\text{odd}}^{\infty} \frac{n-1}{(n-1)!}+\sum_{n=0,\text{odd}}^{\infty} \frac{1}{(n-1)!}$$
$$=S+\sum_{n=0,\text{odd}}^{\infty} \frac{(n-1)}{(n-1)!}$$
$$=S+\sum_{n=3,\text{odd}}^{\infty} \frac{(n-1)}{(n-1)!}$$
$$=S+\sum_{n=3,\text{odd}}^{\infty} \frac{1}{(n-2)!}$$
$$=S+\sum_{n=1,\text{odd}}^{\infty} \frac{1}{n!}$$
$$=S+\sum_{n=0,\text{odd}}^{\infty} \frac{1}{n!}$$
$$=S+\sinh 1$$
$$=\cosh 1+\sinh 1$$

This gives an answer of,
$$\frac{1}{4} \left(3\cosh 1+2\sinh 1 \right)$$
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