Proving that $\sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k=4^{-n}~{2n \choose n}.$ I have happened to have  proved this sum while attempting to prove another summation. Let
$$S_n=\sum_{k=0}^{2n}  {2k \choose k }  {2n \choose k}\left( \frac{-1}{2} \right)^k$$
${2k \choose k} $ is the coefficient of $x^0$ in $(x+1/x)^{2k}$.
Consequently, $S_n$ is the coefficient of $x^0$ in $$f(x)= \sum_{k=0}^{2n} {2n \choose k}~\left (x+\frac{1}{x}\right)^{2k}~\left ( \frac{-1}{2}\right)^{k}=\left(1-\left(\frac{x+1/x}{\sqrt{2}}\right)^2 \right)^{2n} = 4^{-n} ~ \left(x^2+\frac{1}{x^2}\right)^{2n}.$$ Finally, the coefficient of $x^0$ 
in $f(x)$ is $$4^{-n}~{2n \choose n}=S_n.$$
I hope that you will find it interesting and prove it in some other way. Do try!
 A: We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. Recalling the generating function of the central binomial coefficients we can write 
\begin{align*}
[z^n]\frac{1}{\sqrt{1-4z}}=\binom{2n}{n}\tag{1}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^{2n}}&\color{blue}{\binom{2n}{k}\binom{2k}{k}\left(-\frac{1}{2}\right)^k}\\
&=\sum_{k=0}^{2n}\binom{2n}{k}[z^k]\frac{1}{\sqrt{1+2z}}\tag{2}\\
&=[z^0]\frac{1}{\sqrt{1+2z}}\sum_{k=0}^{2n}\binom{2n}{k}z^{-k}\tag{3}\\
&=[z^0]\frac{1}{\sqrt{1+2z}}\left(1+\frac{1}{z}\right)^{2n}\tag{4}\\
&=[z^{-1}]\frac{(1+z)^{2n}}{z^{2n+1}\sqrt{1+2z}}\tag{5}\\
&=[t^{-1}]\frac{\left(1+\frac{t}{1-t}\right)^{2n}}{\left(\frac{t}{1-t}\right)^{2n+1}\sqrt{1+\frac{2t}{1-t}}}\cdot\frac{1}{(1-t)^2}\tag{6}\\
&=[t^{2n}]\frac{1}{\sqrt{1-t^2}}\tag{7}\\
&\,\,\color{blue}{=\frac{1}{4^n}\binom{2n}{n}}\tag{8}
\end{align*}

Comment:


*

*In (2) we apply the coefficient of operator according to (1).

*In (3) we use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (4) we apply the binomial theorem.

*In (5) we write the expression using formal residual by applying again the rule from comment (3).

*In (6) we  use the substitution  $z=\frac{t}{1-t}, dz=\frac{1}{(1-t)^2}dt$.

*In (7) we do some simplifications.

*In (8) we select the coefficient of $t^{2n}$ by taking  (1)  evaluated  at  $z=\frac{1}{4}t^2$.
A: We will use
$$
\frac1{1-x}\left(\frac{x}{1-x}\right)^k=\sum_{n=0}^\infty\binom{n}{k}x^n\tag1
$$
and
$$
(1-4x)^{-1/2}=\sum_{k=0}^\infty\binom{2k}{k}x^k\tag2
$$
Extracting the even part of $(1)$
$$
\begin{align}
\sum_{n=0}^\infty\binom{2n}{k}x^{2n}
=\frac12\left[\frac1{1-x}\left(\frac{x}{1-x}\right)^k+\frac1{1+x}\left(-\frac{x}{1+x}\right)^k\right]\tag3
\end{align}
$$
Compute the generating function of the sum we want
$$
\begin{align}
&\sum_{n=0}^\infty\sum_{k=0}^\infty\binom{2k}{k}\binom{2n}{k}\left(-\frac12\right)^kx^{2n}\tag4\\
&=\frac1{2-2x}\sum_{k=0}^\infty\binom{2k}{k}\left(-\frac{x}{2-2x}\right)^k
+\frac1{2+2x}\sum_{k=0}^\infty\binom{2k}{k}\left(\frac{x}{2+2x}\right)^k\tag5\\
&=\frac1{2-2x}\left(1+\frac{4x}{2-2x}\right)^{-1/2}
+\frac1{2+2x}\left(1-\frac{4x}{2+2x}\right)^{-1/2}\tag6\\
&=\frac1{2-2x}\left(\frac{1-x}{1+x}\right)^{1/2}
+\frac1{2+2x}\left(\frac{1+x}{1-x}\right)^{1/2}\tag7\\[6pt]
&=\left(1-x^2\right)^{-1/2}\tag8\\[6pt]
&=\sum_{n=0}^\infty\frac1{4^n}\binom{2n}{n}x^{2n}\tag9
\end{align}
$$
Explanation:
$(4)$: compute the generating function
$(5)$: apply $(3)$
$(6)$: apply $(2)$
$(7)$: simplify
$(8)$: simplify
$(9)$: apply $(2)$ 
Equating coefficients of $x^{2n}$ gives
$$
\sum_{k=0}^\infty\binom{2k}{k}\binom{2n}{k}\left(-\frac12\right)^k
=\frac1{4^n}\binom{2n}{n}\tag{10}
$$
A: $2kCk=$ coefficient of $x^k$ in $(1+x)^{2k}$ = constant term in $(\frac{(1+x)^2}{x}^){k}=(x+1/x+2)^k$. 
So constant term in $$\sum_{k=0}^{2n}  (x+1/x+2)^k {2n \choose k}\left( \frac{-1}{2} \right)^k$$.
$k \rightarrow 2n-k$ and $2nCk=2nC{2n-k}$ (basically writing sequence in reverse order) 
$\implies$ Constant term in 
$$\sum_{k=0}^{2n}  (x+1/x+2)^{2n-k} {2n \choose k}\left( \frac{-1}{2} \right)^{2n-k}$$
$\implies$ Constant term in 
$$4^{-n} \sum_{k=0}^{2n}  (x+1/x+2)^{2n-k} {2n \choose k}(-2)^{k}$$
By binomial expansion it is constant term in
$$4^{-n} (x+1/x+2-2)^{2n}= 4^{-n}(x+1/x)^{2n} = 4^{-n} \sum_{r=0}^{n} {n \choose r}{n \choose n-r} = 4^{-n} {2n \choose n}$$ by vandermonde's identify.
(the last result follows from selection of the term x, r times (r=0,1,...n) in the expansion implies we need to select to the term 1/x, n-r times for the resulting product to become a constant term).
