Showing $\sum_{k=0}^n \binom{2k}{k} \left(\frac{1}{4}\right)^k=\frac{2n+1}{4^n} \binom{2n}{n} $

Question

I want to show that : \begin{equation} \sum_{k=0}^n \binom{2k}{k} \left(\frac{1}{4}\right)^k=\frac{2n+1}{4^n} \binom{2n}{n} \end{equation} I have no idea how to. I know it's the series for $\frac{1}{\sqrt{1-x}}$ at $1$ which diverges, but that doesn't give much information on the closed form of the sum. If there's a trick to solve this, I would be eager to know it.

Answer 1

I think we can show this by induction. The base case $n=0$ is clear. For the induction step, we have that $$ \sum_{k=0}^{n+1} \binom{2k}{k}\left( \frac{1}{k}\right)^k = \frac{2n+1}{4^n}\binom{2n}{n} + \frac{1}{4^{n+1}} \binom{2(n+1)}{n+1}. $$ To evaluate this sum, recall that $\binom{n}{k}k=\binom{n-1}{k-1}n$ holds for all $n, k$, and so we have $$ \begin{align*} \frac{2n+1}{4^n}\binom{2n}{n} &= \frac{n+1}{4^n} \binom{2n+1}{n+1} \\ &= \frac{n+1}{4^n} \binom{2n+1}{n} \\ &= \frac{2n+2}{2 \cdot 4^n} \binom{2n+1}{n} \\ &= \frac{n+1}{2 \cdot 4^n} \binom{2n+2}{n+1}. \end{align*}$$ This way, we get that $$ \frac{2n+1}{4^n}\binom{2n}{n} + \frac{1}{4^{n+1}} \binom{2(n+1)}{n+1} = \left( \frac{n+1}{2 \cdot 4^n} + \frac{1}{4^{n+1}} \right) \binom{2(n+1)}{n+1} = \frac{2(n+1)+2}{4^{n+1}} \binom{2(n+1)}{n+1}, $$ which shows the inductive step.