finding complex binomial sum in closed form Find sum of $\displaystyle \sum^{n}_{k=0}(-1)^k2^{2k}\binom{n}{k}\binom{2(n-k)}{n-k}$
Attempt: Coefficients of $x^k$ in $(1+x)^k$ and coefficients of $x^{n-k}$ in $(1+x)^{2(n-k)}$ 
could some help me how to solve it, thanks
 A: We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g.
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k2^{2k}\binom{n}{k}\binom{2n-2k}{n-k}}\\
&=\sum_{k=0}^\infty(-1)^k2^{2k}[z^k](1+z)^n[u^{n-k}](1+u)^{2n-2k}\tag{1}\\
&=[u^n](1+u)^{2n}\sum_{k=0}^\infty(-1)^k2^{2k}\frac{u^k}{(1+u)^{2k}}[z^k](1+z)^n\tag{2}\\
&=[u^n](1+u)^{2n}\sum_{k=0}^\infty\left(-\frac{4u}{(1+u)^2}\right)^k[z^k](1+z)^n\tag{3}\\
&=[u^n](1+u)^{2n}\left(1-\frac{4u}{(1+u)^2}\right)^n\tag{4}\\
&=[u^n](1-u)^{2n}\tag{5}\\
&\color{blue}{=(-1)^n\binom{2n}{n}}
\end{align*}

Comment:


*

*In (1) we apply the coefficient of operator twice and we extend the upper limit of the sum to $\infty$ without changing anything, since we are adding zeros only.

*In (2) we use the linearity of the coefficient of operator and apply the rule
\begin{align*}
[z^{p-q}]A(z)=[z^p]z^qA(z)
\end{align*}

*In (3) we apply the substitution rule of the coefficient of operator with $z:=-\frac{4u}{(1+u)^2}$
\begin{align*}
A(u)=\sum_{n=0}^\infty a_nu^n=\sum_{n=0}^\infty u^n[z^n]A(z)
\end{align*}

*In (4) we do some simplifications.

*In (5) we select the coefficient of $u^{n}$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Hereafter, I'll use the
  identity
  $\ds{{2m \choose m} = {-1/2 \choose m}\pars{-4}^{m}}$.

\begin{align}
&\sum_{k=0}^{n}\pars{-1}^{k}\,2^{2k}{n \choose k}{2\bracks{n - k} \choose n - k} =
\sum_{k=0}^{n}\pars{-1}^{k}\,2^{2k}{n \choose k}{-1/2 \choose n - k}
\pars{-4}^{n - k}
\\[5mm] = &\
\pars{-4}^{n}\sum_{k=0}^{n}{n \choose k}{-1/2 \choose n - k} =
\pars{-1}^{n}\,2^{2n}\sum_{k=0}^{n}{n \choose k}
\bracks{z^{n - k}}\pars{1 + z}^{-1/2}
\\[5mm] = &\
\pars{-1}^{n}\,2^{2n}
\bracks{z^{n}}\bracks{\pars{1 + z}^{-1/2}\sum_{k=0}^{n}{n \choose k}z^{k}} =
\pars{-1}^{n}\,2^{2n}\bracks{z^{n}}\pars{1 + z}^{n - 1/2}
\\[5mm] = &\
\pars{-1}^{n}\,2^{2n}{n - 1/2 \choose n} =
\pars{-1}^{n}\,2^{2n}{-1/2 \choose n}\pars{-1}^{n} =
2^{2n}{2n \choose n}\pars{-4}^{-n} =
\bbx{\ds{\pars{-1}^{n}{2n \choose n}}}
\end{align}
A: I would like to contribute to the collection, the challenge being that
we  use a  method that  is different  from the  two that  were already
presented. We seek to evaluate
$$\sum_{k=0}^n (-1)^k 2^{2k} {n\choose k} {2(n-k)\choose n-k}.$$
We use the integral
$${2n-2k\choose n-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} \frac{1}{(1-z)^{n-k+1}}
\; dz.$$
We obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{(1-z)^{n+1}}
\sum_{k=0}^n {n\choose k} (-1)^k 2^{2k} z^k (1-z)^k
\; dz
\\ =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{(1-z)^{n+1}}
(1-4z(1-z))^n
\; dz
\\ =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{(1-z)^{n+1}}
(1-2z)^{2n}
\; dz.$$
This is
$$\sum_{q=0}^n (-1)^q 2^q {2n\choose q} {2n-q\choose n}.$$
We have
$${2n\choose q} {2n-q\choose n} =
\frac{(2n)!}{q! n! (n-q)!} = {2n\choose n} {n\choose q}$$
so this becomes
$${2n\choose n} \sum_{q=0}^n (-1)^q 2^q {n\choose q}
= {2n\choose n} (1-2)^n $$
which is
$$\bbox[5px,border:2px solid #00A000]{
(-1)^n {2n\choose n}.}$$
