Value of the binomial series $\sum_{i=0}^k \frac{{2i \choose i}}{4^i}$ Some time ago a question was asked here regarding the value of the sum $$\sum_{i=0}^k \frac{{2i \choose i}}{4^i}$$. 
But it was deleted later by the OP. I went around it but didn't find a solution.
Some common combinatoric identities I know of are $4^i=2^{2i}=\sum_{j=0}^{2i} {2i\choose j}$ also that ${2i \choose i}=\sum_{j=0}^{i} {i\choose j}^2$. But they were hardly of any use. I also thought of individual term as probability but turns out I can't think of it as anything sensible.a quick WA check gave the answer as $${k+\frac{1}{2} \choose k}=\frac{(2k+1)!}{k!^24^k}$$
Help would be appreciated!

We prove that $$\sum_{k=0}^{n} \frac{{2k \choose k}}{4^k} = \frac{(2n+1)!}{n!^24^n}.$$
We proceed by induction. The base case is easily verifiable. Denote the RHS of the above equation as $a_n$.
Now consider $$a_{n+1} = \frac{(2n+3)!}{(n+1)!^24^{n+1}} = \frac{(2n+2)(2n+3)}{4(n+1)^2} \cdot \frac{(2n+1)!}{n!^24^n} = \frac{(2n+3)}{2n+2} a_n.$$
So we now have $$a_{n+1} - a_n =\frac{a_n}{2n+2} = \frac{(2n+1)!}{2(n+1)\cdot n!^24^n} = \frac{(2n+1)!}{(n+1)!n!4^{n+1}}\cdot 2\\ = \frac{(2n+1)!}{(n+1)!n!4^{n+1}}\cdot \frac{2n+2}{n+1} = \frac{(2n+2)!}{(n+1)!^24^{n+1}} = \frac{{2(n+1) \choose n+1}}{4^{n+1}},$$
which is the added term in the summation. This completes the induction.
