I have to evaluate this expression: $\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}$,
(In the original question we had $\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{k}$)
this is what I have done:
$\begin{aligned} \sum_n\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}x^n & = \sum_k\sum_n\binom{n}{k}\binom{2k}{k}(-2)^{-k}x^n \\ & = \sum_k\binom{2k}{k}(-2)^{-k}\sum_n\binom{n}{k}x^n \\ & = \sum_k\binom{2k}{k}(-2)^{-k}\frac{x^k}{(1-x)^{k+1}} \\ & = \frac{1}{1-x}\sum_k\binom{2k}{k}(\frac{x}{2x-2})^k \end{aligned}$
now we know that $\sum_k\binom{2k}{k}x^k=\frac{1}{\sqrt{1-4x}}$ and so we get
$ \sum_n\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}x^n = \frac{1}{\sqrt{1-x^2}} $
So the sum that I'm looking for is equal to $\sum_{k=0}^n\binom{-1/2}{k}(-1)^k\binom{-1/2}{n-k}$
there is a way to express this expression without using sums or menus signs?