How to simplify $$\sum_{n \geq0 } \binom{2n}{n}\frac{1}{2^n}t^n$$ to show it equals $(1-2t)^{-1/2}$?
Here is a proof I came up with, but I use the fact that I know what the answer should be, so I am a bit dissatisfied with it. By the binomial theorem, the coefficient of $t^n$ in the $(1-2t)^{-1/2}$ is $\binom{-1/2}{n}(-2)^n$, and we want to show this equals $\binom{2n}{n}\frac{1}{2^n}$. This can be done via induction; it amounts to showing $$ 1 \cdot 3 \cdot 5 \cdots (2n-1)\cdot 2^n = (2n)(2n-1)\cdots (n+1)$$ which can be easily proved via induction.
I tried using $\binom{2n}{n}=\sum_{i}\binom{n}{i}\binom{n}{n-i}$
