# How can I simplify this sum?

I have the following sum $$\sum_{a=0}^{[p/2]}\frac{p!}{(a!)^2(p-2a)!}2^{p-2a},$$ where $[p/2]$ denotes the closest integer and $p \in Z^+$.

Is there any way to make the sum look simpler? Especially in regards to the fact that the upper bound bust be an integer.

Any suggestion will be appreciated.

(This is what I found to be the sum of the $2p$th powers of some lengths.)

HINT $$\sum_{a=0}^{[p/2]}\frac{p!}{(a!)^2(p-2a)!}2^{p-2a}=\sum_{a=0}^{[p/2]}\frac{p!(p-a)!}{a!(p-a)!a!(p-2a)!}2^{p-2a}=\sum_{a=0}^{[p/2]}\binom{p}{a}2^{-a}\binom{p-a}{a}2^{p-a}$$
it is the coefficient of $1$ in the expansion of $(2+x+x^{-1})^p$. But $$2+x+x^{-1} = \frac{(x+1)^2}{x}$$
So it is the coefficient of $x^p$ in the expansion of $(x+1)^{2p}$, that is $\binom{2p}{p}$.