TITLE: Non-inductive, not combinatorial proof of $\sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$
QUESTION [7 upvotes]: I've seen the identity $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ used here recently.
I checked for proofs here http://www.proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients
I couldn't have figured out the combinatorial proof by myself, and the inductive proof assumes you already know the answer...
So my question is : do you know how to prove directly through computation that $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ ?

REPLY [8 votes]: $\newcommand{\+}{^{\dagger}}
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${\tt\color{#f00}{\mbox{This method is a "direct calculation" which is very}}}$
${\tt\color{#f00}{\mbox{ convenient when we don't know the answer and, in addition,}}}$
${\tt\color{#f00}{\mbox{ we don't have to guess combinations of Newton binomials.}}}$
It's based in the identity:

$$
{m \choose s}
=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{m} \over z^{s + 1}}\,{\dd z \over 2\pi\ic}
$$

\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 0}^{n}{n \choose k}^{2}} =
\sum_{k = 0}^{n}{n \choose k}
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over
z^{k + 1}}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z}
\sum_{k = 0}^{n}{n \choose k}\pars{1 \over z}^{k}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z}
\pars{1 + {1 \over z}}^{n}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n} \over z^{n + 1}}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\bbox[10px,border:1px groove navy]{2n \choose n}
\end{align}