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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.)

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HINT $$\sum_{a=0}^{[p/2]}\frac{p!}{(a!)^2(p-2a)!}2^{p-2a}=\sum_{a=0}^{[p/2]}\frac{p!(p-a)!}{a!(p-a)!a!(p-2a)!}2^{p-2a}=\sum_{a=0}^{[p/2]}\binom{p}{a}2^{-a}\binom{p-a}{a}2^{p-a}$$

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  • $\begingroup$ Thanks, but this is where from I derived the sum above in the first place. $\endgroup$ – Ryuky Jan 28 '13 at 18:30
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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 $\binom{2p}{p}$.

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