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

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
$$
\sum_{i=0}^n \frac{1}{2^{2i}}\binom{2i}{i}=\frac{1}{2^{2n}}\binom{2n}{n}(2n+1)
$$
