what is the summation of this weird finite sequence? What is the summation of 
$$\sum\limits_{i = 0}^n {\frac{{\left( {\begin{array}{*{20}{c}}
n\\
i
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
n\\
i
\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}
{2n}\\
{2i}
\end{array}} \right)}}}$$
here $n$ is a constant.
another problem from my textbook. Another weird problem:( 
Any one could help?
 A: Examination of the numerical data reveals the pattern
$$
\sum_{i=0}^n \frac{\binom ni^2}{\binom{2n}{2i}} = \frac{4^n}{\binom{2n}n}.
$$
To prove this, multiply the given sum by $\binom{2n}n$ and simplify the factorials:
\begin{align*}
\binom{2n}n \sum_{i=0}^n \frac{\binom ni^2}{\binom{2n}{2i}} &= \frac{(2n)!}{n!^2} \sum_{i=0}^n \bigg( \frac{n!}{i!(n-i)!} \bigg)^2 \bigg/ \frac{(2n)!}{(2i)!(2n-2i)!} \\
&= \sum_{i=0}^n \frac{(2i)!(2n-2i)!}{i!^2(n-i)!^2} \\
&= \sum_{i=0}^n \binom{2i}i \binom{2(n-i)}{n-i}.
\end{align*}
This is the $n$th element of the convolution of the sequence $\{\binom{2m}m\}$ with itself, therefore its generating function is the square of the function $\sum_{m=0}^\infty \{\binom{2m}m\} x^m = 1/\sqrt{1-4x}$. In other words,
$$
\sum_{n=0}^\infty x^n \binom{2n}n \sum_{i=0}^n \frac{\binom ni^2}{\binom{2n}{2i}} = \bigg( \frac1{\sqrt{1-4x}} \bigg)^2 = \frac1{1-4x} = \sum_{n=0}^\infty (4x)^n;
$$
comparing the coefficients of $x^n$ on both sides, we see that $\binom{2n}n \sum_{i=0}^n {\binom ni^2}\big/{\binom{2n}{2i}} = 4^n$ as desired.
A: I'll work with the summand.
$\begin{array}\\
\frac{\binom{n}{i}^2}{\binom{2n}{2i}}
&=\frac{\big(\frac{n!}{i!(n-i)!}\big)^2}{\frac{(2n)!}{(2i)!(2n-2i)!}}\\
&=\frac{\frac{(2i)!(2n-2i)!}{(i!(n-i)!)^2}}{\frac{(2n)!}{n!^2}}\\
&=\frac{\binom{2i}{i}\binom{2n-2i}{n-i}}{\binom{2n}{n}}\\
\end{array}
$
Therefore
$\sum_{i=0}^n \frac{\binom{n}{i}^2}{\binom{2n}{2i}}
=\sum_{i=0}^n \frac{\binom{2i}{i}\binom{2n-2i}{n-i}}{\binom{2n}{n}}
=\frac1{\binom{2n}{n}}\sum_{i=0}^n \binom{2i}{i}\binom{2n-2i}{n-i}
$.
If
$G(x)
=\sum_{i=0}^{\infty} g_i x^i
$,
$\begin{array}\\
G^2(x)
&=\sum_{i=0}^{\infty} \sum_{j=0}^{\infty} g_i  g_j x^{i+j}\\
&=\sum_{m=0}^{\infty} \sum_{k=0}^m g_k  g_{m-k} x^{m}\\
&=\sum_{m=0}^{\infty} x^{m}\sum_{k=0}^m g_k  g_{m-k} \\
\end{array}
$
Therefore,
if
$T(x)
=\sum_{i=0}^{\infty} \binom{2i}{i}x^i
$,
$T^2(x)
=\sum_{n=0}^{\infty} x^{n}\sum_{i=0}^n \binom{2i}{i} \binom{2n-2i}{n-i}
$.
$T(x)$ is well known -
find it and square it.
