I'm trying to solve the closed form for this:
$\sum_k\left(\begin{array}{l}n \\ k\end{array}\right)\left(\begin{array}{c}2 k \\ k\end{array}\right)\left(-\frac{1}{2}\right)^k$
Meanwhile the answer for this has been given to us in the book as:
5.70 This is $F\left(-n, \frac{1}{2} ; 1 ; 2\right)$; but it's also $(-2)^{-n}\left(\begin{array}{c}2 n \\ n\end{array}\right) F\left(-n,-n ; \frac{1}{2}-n ; \frac{1}{2}\right)$ if we replace $k$ by $n-k$. Now $F\left(-n,-n ; \frac{1}{2}-n ; \frac{1}{2}\right)=F\left(-\frac{n}{2},-\frac{n}{2} ; \frac{1}{2}-n ; 1\right)$ by Gauss's identity (5.111). (Alternatively, $F\left(-n,-n ; \frac{1}{2}-n ; \frac{1}{2}\right)=2^{-n} F\left(-n, \frac{1}{2} ; \frac{1}{2}-n ;-1\right)$ by the reflection law (5.101), and Kummer's formula (5.94) relates this to (5.55).) The answer is 0 when $n$ is odd, $2^{-n}\left(\begin{array}{c}n \\ n / 2\end{array}\right)$ when $n$ is even. (See [164, $\S 1.2]$ for another derivation. This sum arises in the study of a simple search algorithm [195].)
I don't understand , could someone help me understand this.