I want to prove that the $2n$ Hermite polynomial is $H_{2n}(x)=(-1)^n\sum_{s=0}^{n}(-1)^{2s}(2x)^{2s}\cfrac{(2n)!}{(2s)!(n-s)!}$
So I resort to equality:
$\begin{equation*} H_n(x)=\sum_{s=0}^{n/2}(-1)^s\cfrac{n!}{(n-2s)!s!}(sx)^{n-2s} \end{equation*}$
So if I substitute $2n$ in place of $n$ I get to that equality becomes:
$\begin{align*} H_n(x)&=\sum_{s=0}^{n/2}(-1)^s\cfrac{n!}{(n-2s)!s!}(sx)^{n-2s}\\ H_{2n}(x)&=\sum_{s=0}^{2n/2}(-1)^s\cfrac{(2n)!}{(2n-2s)!s!}(sx)^{2n-2s}\\ H_{2n}(x)&=\sum_{s=0}^{n}(-1)^s\cfrac{(2n)!}{(2n-2s)!s!}(sx)^{2n-2s}\\ \end{align*}$
But then I don't know what else I can do to get the desired result (the beginning)
Or if maybe I have to use the following method:
Since $H_n(x)=(-1)^ne^{x^2}\cfrac{d^n}{dx^n}e^{-x^2}$ then:
$\begin{align*} H_{2n}(x)&=(-1)^{2n}e^{x^2}\cfrac{d^{2n}}{dx^{2n}}e^{-x^2}\\ H_{2n}(x)&=\left[(-1)^2\right]^ne^{x^2}\cfrac{d^{2n}}{dx^{2n}}e^{-x^2}\\ H_{2n}(x)&=1^ne^{x^2}\cfrac{d^{2n}}{dx^{2n}}e^{-x^2}\\ H_{2n}(x)&=e^{x^2}\cfrac{d^{2n}}{dx^{2n}}e^{-x^2}\\ \end{align*}$
In which case I don't know how to find the desired result.
I really appreciate your help to know how to get to the result $(-1)^n\sum_{s=0}^{n}(-1)^{2s}(2x)^{2s}\cfrac{(2n)!}{(2s)!(n-s)!}$