Non-inductive, not combinatorial proof of $\sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ 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 $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ ?
 A: lab bhattacharjee has given the proof, but it is worth pointing out that this formula is simply an application of the Vandermonde convolution, which says that
$$\sum_{j=0}^k \binom{n}{j}\binom{m}{k-j} = \binom{n+m}{k}.$$
Setting $n=m=k$ and noting that $\binom{n}{n-j}=\binom{n}{j}$ gives the result. 
EDIT: By the way, for a slightly non-standard (but purely technical, as you require) proof of the Vandermonde convolution, let $X_1,\cdots,X_{n+m}$ be independent Bernoulli distributed RV's with parameter $p=1/2$. Then, $S_{n+m}=\sum_{j=0}^{n+m}X_j$ has binomial distribution $P[S_{n+m}=k]=\binom{n+m}{k}(\frac{1}{2})^{n+m}$. Applying the ordinary discrete convolution formula for probability distributions yields
$$(1/2)^{n+m}\binom{n+m}{k}=P[S_{n+m}=k]=\sum_{j=0}^k P[S_n=j]P[S_m=k-j]=(1/2)^{n+m}\sum_{j=0}^k\binom{n}{j}\binom{m}{k-j}.$$
A: $\newcommand{\+}{^{\dagger}}
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${\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}
A: Consider the identity $(1+x)^{2n}=(1+x)^n\cdot(1+x)^n$.  By the binomial theorem we have $\displaystyle(1+x)^n=\sum_{k=0}^n{n\choose k} x^k$, so multiplying out we compute the right hand side as $\displaystyle(1+x)^n\cdot(1+x)^n = \sum_{k=0}^{2n}\left(\sum_{j=0}^k{n\choose j}{n\choose k-j}x^k\right)$.  But the LHS is just $\displaystyle(1+x)^{2n} = \sum_{k=0}^{2n}{2n\choose k}x^k$; equating coefficients of $x^n$ we get $\displaystyle{2n\choose n}=\sum_{j=0}^n {n\choose j}{n\choose n-j}$.  Finally, using the identity ${n\choose j}={n\choose n-j}$ gives the desired result.
A: You're after a sum of a hypergeometric term. There are general techniques for finding closed forms or proving that they don't exist. See e.g. the book A = B, or MathWorld's description of Sister Celine's method, Gosper's algorithm, and Zeilberger's algorithm.
A: Note that we can divide the $2n$ objects in $2$ groups including $n$ elements each.
Then to choose $n$ elements out of $2n$ we can choose $0$ objects in the first group and $n$ in the second, or $1$ in the first and $n-1$ in the second and so on, i.e $$\binom{2n}{n} = \sum_{i = 0}^{2n}\binom{n}{i}\binom{n}{n-i} =\sum_{i = 0}^{2n}{\binom{n}{i}}^2 $$
