Sum of binomial coefficients with powered variable I am working with two terms like this
$$\sum_{i=0}^n {2i \choose i} {2n-2i \choose n-i}, \quad \sum_{i=0}^n {2n - 2i \choose n-i} 4^i$$ 
The latter will be
$$\sum_{i=0}^n {2n - 2i \choose n-i} 4^i = (2n+1) {2n \choose n}$$
I wonder how to prove such a result.
Thank you.
 A: The generating function of the central binomial coefficients $\binom{2n}n$ is $\frac1{\sqrt{1-4x}}$ (see Generating functions and central binomial coefficient). We have
\begin{eqnarray}
\sum_{n=0}^\infty\sum_{i=0}^n\binom{2n-2i}{n-i}4^ix^n
&=&
\sum_{i=0}^\infty\sum_{n=i}^\infty\binom{2n-2i}{n-i}4^ix^n
\\
&=&
\sum_{i=0}^\infty4^ix^i\sum_{n=i}^\infty\binom{2n-2i}{n-i}x^{n-i}
\\
&=&
\sum_{i=0}^\infty4^ix^i\sum_{n=0}^\infty\binom{2n}nx^n
\\
&=&
\frac1{1-4x}\cdot\frac1{\sqrt{1-4x}}
\\
&=&
(1-4x)^{-\frac32}
\\
&=&
\frac12\frac{\mathrm d}{\mathrm dx}\frac1{\sqrt{1-4x}}
\\
&=&
\frac12\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty\binom{2n}nx^n
\\
&=&
\frac12\sum_{n=1}^\infty\binom{2n}nnx^{n-1}
\\
&=&
\frac12\sum_{n=0}^\infty\binom{2n+2}{n+1}(n+1)x^n
\\
&=&
\sum_{n=0}^\infty\binom{2n}n(2n+1)x^n\;.
\end{eqnarray}
A: The first one is easy to compute with generating functions: it is the convolution of $\binom{2n}{n}$ with itself, hence its generating function is
$$\left(\sum_{n=0}^\infty\binom{2n}{n}x^n\right)^2 = \frac{1}{1-4x}$$
so $\sum_{i=0}^n\binom{2i}{i}\binom{2(n-i)}{n-i} = 4^n$.
A combinatorial argument is suprisingly hard, see
Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$
