<s>[INST] Generating fuctions: $\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k=2^{-2n}\binom{2n}{n}$ I'm trying to solve one of my combinatorics exercise but I struggle a bit.

Is the equality correct for all the $n\ge 0$? 
  $$\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k=2^{-2n}\binom{2n}{n}$$

First of all:$$\sum_{n=0}^\infty \left(2^{-2n}\binom{2n}{n}\right)x^n=\sum_{n=0}^\infty \left( \frac{x}{4}\right)^n\binom{2n}{n}=\frac{1}{\sqrt{1-x}}$$
Now
$$\sum_{n=0}^\infty \left(\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k \right)x^n=\sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k\sum_{n=k}^\infty \binom{2k}{k}\binom{n}{k} x^n=[n-k=m]=\sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k\sum_{m=0}^\infty \binom{2k}{k}\binom{m+k}{m} x^{m+k}=\sum_{k=0}^\infty \left(-\frac{x}{4}\right)^k\binom{2k}{k}\sum_{m=0}^\infty \binom{m+k}{m} x^{m}$$
and here I don't know what to do next. Can anyone help me? Thanks in advice!
 [/INST] This follows directly (more or less) from Vandermonde's identity:
\begin{align*}
  \sum_{k=0}^n \binom{n}{k}\binom{-1/2}{k} &= \binom{n-1/2}{n} \\
       &= \frac{(n-1/2)(n-3/2)\cdots(n-1/2-(n-1))}{n!}\\
       &= \frac{(2n-1)(2n-3)\cdots (1)}{2^nn!} \\
       &= 2^{-2n}\binom{2n}{n}.
\end{align*}
But also,
$$\binom{-1/2}{k} = \frac{(-1/2)(-1/2-1)\cdots(-1/2-(k-1)}{k!}
      = (-1)^k\frac{1\cdot 3\cdot 5\cdots (2k-1)}{2^kk!}
      = (-1)^k 2^{-2k}\binom{2k}{k}$$
and therefore
$$\sum_{k=0}^n \binom{n}{k}\binom{-1/2}{k} = \sum_{k=0}^n \binom{n}{k}(-1)^k2^{-2k}\binom{2k}{k}.$$
</s>