How to find the sum of this expression?

Question

I want to find the sum of the following expression in terms of $n$. I tried looking for it but couldn't come up with one. Any help will be greatly appreciated.

$$\sum_{i=0}^n \frac{1}{4^i}\binom{2i}{i}$$

Answer 1

By the extended binomial theorem we have $$ \sum_{n\geq 0}\frac{x^n}{4^n}\binom{2n}{n}=\frac{1}{\sqrt{1-x}} $$ hence $\sum_{i=0}^{n}\frac{1}{4^i}\binom{2i}{i}$ can be seen as the coefficient of $x^n$ in $\frac{1}{(1-x)\sqrt{1-x}}$, since $$ f(x)=\sum_{n\geq 0}a_n x^n\quad \Longrightarrow\quad \frac{f(x)}{1-x}=\sum_{n\geq 0}A_n x^n,\quad A_n=a_0+a_1+\ldots+a_n.$$ On the other hand $\frac{1}{(1-x)\sqrt{1-x}}$ is just twice the derivative of $\frac{1}{\sqrt{1-x}}$, hence $$ \frac{1}{(1-x)\sqrt{1-x}} = \sum_{n\geq 1}\frac{2n x^{n-1}}{4^n}\binom{2n}{n}$$ and by reindexing we get: $$ \sum_{i=0}^{n}\frac{1}{4^i}\binom{2i}{i} = \frac{2(n+1)}{4^{n+1}}\binom{2n+2}{n+1}=\color{red}{\frac{2n+1}{4^n}\binom{2n}{n}}.$$