How to prove the following binomial identity How to prove that 
$$\sum_{i=0}^n \binom{2i}{i} \left(\frac{1}{2}\right)^{2i} = (2n+1) \binom{2n}{n} \left(\frac{1}{2}\right)^{2n} $$
 A: OK, the correct thing is:   
$$\sum_{i=0}^n \binom{2i}{i} \left(\frac{1}{2}\right)^{2i} = (2n+1) \binom{2n}{n} \left(\frac{1}{2}\right)^{2n}  $$   
1) For $n=0,1$ it's true.
2) Denote the RHS by A(n).
Add the LHS's $i=n+1$ term $T$ to $A(n)$,
and prove that what you get equals $A(n+1)$.
This can be proved directly by simplifying the expression.
This completes your induction.
Side note: here $T = \binom{2n+2}{n+1} \left(\frac{1}{2}\right)^{2n+2}$
A: 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.
A: Let
 $$\begin{align}
f(i)&=(2i+1)\binom{2i}i \left(\frac 12\right)^{2i}\\
\Rightarrow f(i-1)&=\color{purple}{(2i-1)}\color{blue}{\binom{2i-2}{i-1}}\color{green}{\left(\frac 12\right)^{2i-2}}\\
&=\color{purple}{(2i-1)}\cdot\frac{\color{lightblue}{2i(2i-1)}}{\color{grey}{2i(2i-1)}}\cdot \frac {\color{grey}i}{\color{lightblue}i}\cdot \color{blue}{\frac {(2i-2)(2i-3)(2i-4)\cdots (i+1)i}{(i-1)!}}\cdot \color{green}{ 4\left(\frac 12\right)^{2i}}\\
&=\color{purple}{(2i-1)}\cdot\color{blue}{\frac {i\cdot i}{2i(2i-1)}\binom{2i}{i}}\cdot\color{green}{4\left(\frac 12\right)^{2i}}\\
&=2i\color{blue}{\binom {2i}i}\color{green}{\left(\frac 12\right)^{2i}}\\
f(i)-f(i-1)&=(2i+1)\binom{2i}i \left(\frac 12\right)^{2i}-2i\binom {2i}i\left(\frac 12\right)^{2i}\\
&=\binom {2i}i\left(\frac 12\right)^{2i}
\end{align}$$
Hence
$$\begin{align}
\sum_{i=0}^n \binom {2i}i\left(\frac 12\right)^{2i}
&=\sum_{i=1}^n \binom {2i}i\left(\frac 12\right)^{2i}\\
&=\sum_{i=1}^n f(i)-f(i-1)\\
&=f(n)-f(0)\qquad\qquad\text {note that }f(0)=0\\
&=(2n+1)\binom{2n}n\left(\frac 12\right)^{2n}\qquad\blacksquare
\end{align}$$
$$\begin{align}
\end{align}$$
A: Let be $$b_i=\frac{1}{2^{2i}}\binom{2i}{i}\quad \text{and}\quad a_i=\frac{i}{2^{2i-1}}\binom{2i}{i}.$$
Observing that $$a_i=2ib_i$$ and $$a_{i+1}=\frac{i+1}{2^{2i+1}}\binom{2i+2}{i+1}=\frac{i+1}{2}\frac{(2i+2)(2i+1)}{(i+1)(i+1)}\frac{1}{2^{2i}}\binom{2i}{i}=(2i+1)b_i$$
we have
$$
\begin{align*}
a_{i+1}-a_i&=(2i+1)b_i-2ib_i=b_i
\end{align*}
$$
and finally we have a telescoping sum:
$$
\sum_{i=0}^n b_i=\sum_{i=0}^n (a_{i+1}-a_i)=a_{n+1}-a_0=(2n+1)b_n
$$
that is
$$
\sum_{i=0}^n \frac{1}{2^{2i}}\binom{2i}{i}=\frac{1}{2^{2n}}\binom{2n}{n}(2n+1)
$$
