How to prove the following binomial identity How to prove that 
$$\sum_{i=0}^n \binom{2i}{i} \left(\frac{1}{2}\right)^{2i} = (2n+1) \binom{2n}{n} \left(\frac{1}{2}\right)^{2n} $$

Suppose we seek to verify that
$$\sum_{q=0}^n {2q\choose q} 4^{-q} =
(2n+1) {2n\choose n} 4^{-n}$$
using a method other than induction.
Introduce the Iverson bracket
$$[[0\le q\le n]]
= \frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{w^q}{w^{n+1}} \frac{1}{1-w}
\; dw$$
This yields  for the sum  (we extend the  sum to infinity  because the
Iverson bracket controls the range)
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{1}{w^{n+1}} \frac{1}{1-w}
\sum_{q\ge 0} {2q\choose q} w^q 4^{-q}
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{\sqrt{1-w}}
\; dw.$$
We have used the Newton binomial to obtain the square root.
Continuing we have
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{1}{w^{n+1}} \frac{1}{(1-w)^{3/2}}
\; dw.$$
Use the Newton binomial again to obtain
$${n+1/2\choose n}
= \frac{1}{n!} \prod_{q=0}^{n-1} (n+1/2-q)
= \frac{1}{2^n \times n!} \prod_{q=0}^{n-1} (2n+1-2q)
\\ = \frac{1}{2^n \times n!} \frac{(2n+1)!}{2^n n!}
= 4^{-n} \frac{(2n+1)!}{n!\times n!}
= 4^{-n} (2n+1) {2n\choose n}.$$
This is the claim.
