Show that $\sum\limits_{k=0}^n\binom{2n}{2k}^{\!2}-\sum\limits_{k=0}^{n-1}\binom{2n}{2k+1}^{\!2}=(-1)^n\binom{2n}{n}$ How can I prove the identity: 
$$
\sum_{k=0}^n\binom{2n}{2k}^2-\sum_{k=0}^{n-1}\binom{2n}{2k+1}^2=(-1)^n\binom{2n}{n}?
$$
Maybe, can we expand
$$
f(x)=(1+x)^{2n}?
$$
Thank you.
 A: This is simpler if you rewrite your left hand side as
$$
  \sum_{k=0}^n\binom{2n}{2k}^2-\sum_{k=0}^{n-1}\binom{2n}{2k+1}^2=
  \sum_{i=0}^{2n}(-1)^i\binom{2n}i^2=
  \sum_{i=0}^{2n}(-1)^i\binom{2n}i\binom{2n}{2n-i}
$$
first. So you are looking for the coefficient of $X^{2n}$ in the product $(1-X)^{2n}(1+X)^{2n}=(1-X^2)^{2n}$, which is the same as the coefficient of $Y^n$ in $(1-Y)^{2n}$. That coefficient is clearly $(-1)^n\binom{2n}n$.
Note that one could replace $2n$ by $m$ and allow it to be odd, if one takes the right hand side to be$~0$ in that case. The proof remains the same, remarking that there is obviously no term $X^m$ in $(1-X^2)^m$ when $m$ is odd. 
Let me also give a combinatorial proof (since that tag is given). Given $m$ mixed-sex couples, the left hand side counts the number of gender-balanced subsets (containing equally many women as men), counted with a factor$~{-}1$ if that number for each sex is odd. In this counting, the subsets that do not contain exactly one member of each couple cancel out as follows: for such a subset, find the first couple of which which not exactly one member was selected, and add or remove the couple to obtain a cancelling subset; this operation is clearly a gender parity respecting involution. So we are left with as un-cancelled contributions those from the gender-balanced subsets with one member from each couple, which subsets are clearly of size$~m$. If $m$ is odd then no subsets at all remain, as gender balance is impossible. If $m$ is even then each remaining gender-balanced subset is counted with sign $(-1)^{m/2}$and it is determined by the $m/2$ women it contains (it is completed by the men from the other $m/2~$couples), so there are $\binom m{m/2}$ of them.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\sum_{k = 0}^{n}{2n \choose 2k}^{2}
     -\sum_{k = 0}^{n - 1}{2n \choose 2k + 1}^{2} = \pars{-1}^{n}{2n \choose n}:
     \ {\large ?}}$.

$$
\mbox{Note that}\quad
\sum_{k = 0}^{n}{2n \choose 2k}^{2} - \sum_{k = 0}^{n - 1}{2n \choose 2k + 1}^{2}
=\sum_{k = 0}^{2n}\pars{-1}^{k}{2n \choose k}^{2}
$$

\begin{align}&\color{#66f}{\large\sum_{k = 0}^{n}{2n \choose 2k}^{2}
-\sum_{k = 0}^{n - 1}{2n \choose 2k + 1}^{2}}
=\sum_{k = 0}^{2n}{2n \choose k}\pars{-1}^{k}
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}
\\[6mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n} \over z}
\sum_{k = 0}^{2n}{2n \choose k}\pars{-\,{1 \over z}}^{k}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n} \over z}
\,\bracks{1 + \pars{-\,{1 \over z}}}^{2n}\,{\dd z \over 2\pi\ic}
\\[6mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 - z^{2}}^{2n} \over z^{2n + 1}}
\,{\dd z \over 2\pi\ic}
=\sum_{k = 0}^{2n}{2n \choose k}\pars{-1}^{k}\
\underbrace{\oint_{\verts{z}\ =\ 1}{1 \over z^{2n - 2k + 1}}%
\,{\dd z \over 2\pi\ic}}_{\ds{=\ \color{#c00000}{\large\delta_{kn}}}}
\\[6mm]&=\sum_{k = 0}^{2n}{2n \choose k}\pars{-1}^{k}\,\delta_{kn}
=\color{#66f}{\large\pars{-1}^{n}{2n \choose n}}
\end{align}
A: This is simply
$$\sum_{k=0}^{2n} {2n\choose k}^2 (-1)^k
=\sum_{k=0}^{2n} {2n\choose k} {2n\choose 2n-k} (-1)^k
\\ = [z^{2n}] (1+z)^{2n} \sum_{k=0}^{2n} {2n\choose k} z^k (-1)^k
\\ = [z^{2n}] (1+z)^{2n} (1-z)^{2n}
= [z^{2n}] (1-z^2)^{2n} = [z^n] (1-z)^{2n} \\ = (-1)^n {2n\choose n}.$$
A: Idea. We shall obtain the identity by equating the coefficient of $x^{2n}$ in the expansions of $(1+x^2)^{2n}$ and $(1+ix)^{2n}(1-ix)^{2n}$.
The binomial expansion of $(1+x^2)^{2n}$ is
$$
(1+x^2)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}x^{2k},
$$
while
$$
(1+x^2)^{2n}=(1+ix)^{2n}(1-ix)^{2n}=\left(\sum_{k=0}^{2n}\binom{2n}{k}(ix)^{k}\right)\left(\sum_{k=0}^{2n}\binom{2n}{k}(-ix)^{k}\right).
$$
The coefficient of $x^{2n}$ in the first expansion is $a_{2n}=\displaystyle\binom{2n}{n}$, while in the right-hand side of the above it can be expressed as the following sum:
\begin{align}
a_{2n}&=\sum_{k=0}^{2n}i^k(-i)^{2n-k}\binom{2n}{k}\binom{2n}{2n-k}=i^{2n}
\sum_{k=0}^{2n}(-1)^{k}\binom{2n}{k}\binom{2n}{k} \\&=(-1)^{n}
\sum_{k=0}^{2n}(-1)^{k}\binom{2n}{k}^{\!2}=
(-1)^{n}\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}^{\!2}\\ &=
(-1)^{n}\sum_{k=0}^{n}\binom{2n}{2k}^{\!2}-(-1)^{n}\sum_{k=0}^{n-1}\binom{2n}{2k+1}^{\!2}\!\!.
\end{align}
Hence
$$
\sum_{k=0}^n\binom{2n}{2k}^2-\sum_{k=0}^{n-1}\binom{2n}{2k+1}^2=(-1)^n\binom{2n}{n}.
$$
