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!
 A: You may consider that
$$ \frac{1}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{\pi/2}\sin^{2k}(\theta)\,d\theta \tag{A}$$
from which it follows that:
$$ \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{\pi/2}\sum_{k=0}^{n}\binom{n}{k}(-\sin^2\theta)^k =\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta\tag{B}$$
and the conclusion is straightforward through the substitution $\theta\mapsto\frac{\pi}{2}-\varphi$.

With generating functions, from 
$$ \sum_{k\geq 0}\binom{2k}{k}\frac{z^k}{4^k}=\frac{1}{\sqrt{1-z}} \tag{C}$$
by replacing $z$ with $\frac{x}{1+x}$, then by multiplying both sides by $\frac{1}{1+x}$, we get
$$ \sum_{k\geq 0}\binom{2k}{k}\frac{x^k}{4^k(1+x)^{k+1}} = \frac{1}{\sqrt{1-x}}\tag{D}$$
then by applying $[x^n]$ to both sides:
$$ \sum_{k\geq 0}\binom{2k}{k}\frac{(-1)^k}{4^k}\binom{n}{k} = \frac{1}{4^n}\binom{2n}{n}\tag{E}$$
where the RHS has been managed through $(C)$ and the LHS through stars and bars.
A: 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
$$\sum_{k=0}^n \binom{n}{k}\binom{-1/2}{k} = \sum_{k=0}^n \binom{n}{k}(-1)^k2^{-2k}\binom{2k}{k}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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In 0ne of my previous answers I already showed that
  \begin{equation}
{2m \choose m} = {-1/2 \choose m}\pars{-4}^{m}\label{1}\tag{1}
\end{equation}

\begin{align}
\mbox{So,}\quad\sum_{k = 0}^{\infty}{2k \choose k}{n \choose k}\pars{-\,{1 \over 4}}^{k} & =
\sum_{k = 0}^{\infty}{n \choose k}\pars{-\,{1 \over 4}}^{k}\
\overbrace{\bracks{{-1/2 \choose k}\pars{-4}^{k}}}^{\ds{2k \choose k}}
\\[5mm] & =
\sum_{k = 0}^{\infty}{n \choose k}\
\overbrace{\bracks{z^{k}}\pars{1 + z}^{-1/2}}^{\ds{-1/2 \choose k}}\ =\
\bracks{z^{0}}\pars{1 + z}^{-1/2}\sum_{k = 0}^{\infty}{n \choose k}
\pars{1 \over z}^{k}
\\[5mm] & =
\bracks{z^{0}}\pars{1 + z}^{-1/2}\bracks{\pars{1 + {1 \over z}}^{n}} =
\bracks{z^{n}}\pars{1 + z}^{n - 1/2}
\\[5mm] & =
{n - 1/2 \choose n} = {-1/2 \choose n}\pars{-1}^{n} =
\pars{1 \over 4}^{n}\
\overbrace{\bracks{{-1/2 \choose n}\pars{-4}^{n}}}^{\ds{{2n \choose n}.\
\mbox{See}\ \eqref{1}}}
\\[5mm] & =
\bbx{2^{-2n}{2n \choose n}}
\end{align}
