TITLE: what is the summation of such a finite sequence?
QUESTION [5 upvotes]: The summation is: $$\sum_{i=0}^n \binom{2i}i \binom{2n-2i}{n-i}$$
The answer is $4^n$.
How to prove it, and how to think out it?

REPLY [3 votes]: Hint. This sum may be obtained by a Cauchy product. 
Recall that
$$
\sum_{i=0}^n \binom{2i}i x^i=\frac{1}{\sqrt{1-4x}},\quad |x|<\frac14. \tag1
$$
(for a proof, you may use the generalized binomial theorem).
Then, by the Cauchy product:
$$
\begin{align}
\left(\sum_{n=0}^{\infty} \binom{2n}n x^n\right)^2
&=\sum_{n=0}^{\infty}\left( \sum_{i=0}^n \binom{2i}i \binom{2n-2i}{n-i}\right)x^n
\end{align}, \quad |x|<\frac14 \tag2
$$
on the other hand, using $(1)$:
$$
\begin{align}
\left(\sum_{n=0}^{\infty} \binom{2n}n x^n\right)^2 =\frac{1}{1-4x}=\sum_{n=0}^{\infty} 4^n x^n  ,\quad |x|<\frac14, \tag3
\end{align}
$$
you just identify the coefficients in $(2)$ and $(3)$ to obtain your identity.