The following result is Theorem 5.3.1 of Grimmett and Stirzaker (3e.)
Let $0<p<1$ and $q = 1-p$. Then $$\sum_{n\geq 0}\binom{2n}{n}(pqs^2)^n = (1-4pqs^2)^{-1/2}$$
, but I have no idea how to reach this conclusion (the authors do not explain why.)
This power / generating series arises as follows. Put $S_n = \sum_1^n X_i$ where the $X_i$ are i.i.d with $P(X_i = 1) = p$ and $P(X_i = -1) = q$. We consider $P(S_n = 0) = \binom{n}{n/2}(pq)^{n/2}$ if $n$ is even, and 0 otherwise. We encode it into a power series, that is, we have $$\sum_{n\geq 0}P(S_n = 0)s^n = \sum_{n\in 2\mathbb N_{\geq 0}}\binom{n}{n/2}(pq)^{n/2}s^n = \sum_{n\geq 0}\binom{2n}{n}(pqs^2)^n$$
