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!
[Math] Proving that $\sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k=4^{-n}~{2n \choose n}.$
binomial theorembinomial-coefficientssequences-and-series
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Best Answer
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*}
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 $z=\frac{1}{4}t^2$.