A combinatorial identity involving square of central binomial coefficient. While solving a problem  I came across the following interesting identity, valid by numerical evidence:
$$
S_n:=\sum_{k=0}^n\left (-\frac14\right)^k\binom {2k}k^2\frac1 {1-2k}\binom {n+k-2}{2k-2}=\begin {cases}\displaystyle 
\left [ \left (\frac14\right)^m\binom {2m}m\frac1 {1-2m}\right]^2,& n=2m;\\
0,& n=2m+1.
\end{cases}\tag1 $$
Is there a simple way to prove it?

From WA I know:
$$S_n=\frac {(1-(-1)^n)\Gamma^2 (\frac {n-1}2)}{8\pi\Gamma^2 (\frac {n+2}2)}\tag2 $$
Obviously (2) evaluates to 0 for odd $n $. For $n=2m$ the expression gives
$$S_{2m}=\frac {\Gamma^2 (m-\frac12)}{4\pi\Gamma^2 (m+1)}=\frac {\left [\frac {(2m-2)!}{(m-1)!}\frac{\sqrt\pi}{4^{m-1}}\right]^2}{4\pi (m!)^2}\\
=4\left [\frac1 {4^{m}}\frac{m} {(2m)(2m-1)}\frac {(2m)!}{m!m!}\right]^2
=\left [\frac1 {4^{m}}\frac{1} {2m-1}\binom {2m}{m}\right]^2,$$
also in agreement with (1).
However I still wonder how the result can be obtained without resorting to computer help.
 A: In general, definite sums of binomial expressions (technically, hypergeometric terms) can be tackled by a technique called Zeilberger's algorithm. See the book $A = B$ by Petkovšek, Wilf, and Zeilberger. Used to be available legally online in PDF, and maybe still is somewhere.
The actual algorithm is complicated enough that it's better to implement it in a CAS than work it through by hand except in really trivial cases, but knowing its existence allows you to throw the sum at a CAS which implements it. In particular, Wolfram Alpha gave me
$$\sum_{k=0}^{2m+1} \frac{1}{(-4)^k (1 - 2k)} \binom{2k}{k}^2 \binom{2m+1+k-2}{2k-2} = \frac{\pi}{4\Gamma(1-m)^2 \Gamma\left(m+\frac32\right)^2}$$
which can be un-substituted to give
$$\sum_k \frac{1}{(-4)^k(1-2k)} \binom{2k}k^2 \binom{n+k-2}{2k-2} = \frac{\pi}{4\Gamma\left(\frac{3-n}2\right)^2 \Gamma\left(\frac{n+2}{2}\right)^2}$$
which looks like it's a good way towards your target. It also points up an exception to your case analysis: when $n=1$ the sum is non-zero.
