I want to prove the identity $\sum_{i=0}^n{n + i \choose i}\frac{1}{2^i} = 2^n$ combinatorially. I tried multiplying both sides of the equation by $2^n$ and obtained $\sum_{i=0}^n{n + 1 \choose i}2^{n - i} = 2^{2n}$. Whenever I see $2^n$ I think of counting subsets and bitstrings. Here's my attempt to solve the problem.
I will start from the left side of the equation. We can view it as counting bitstrings of length $2n + 1$ where at least $n + 1$ of the bits are $1$. We will count these strings by fixing the index $n + 1 + i$ of the $n + 1$-th one in the string. Note that ${n + i \choose i} = {n + i \choose n}$. So for a fixed $i$, all we have to do is pick the indices of the first $n$ ones and then multiply that by the possibilities of all the remaining bits which is $2^{n - i}$.
My problem is that I don't know how to show that the number of bitstrings of length $2n + 1$ with at least $n + 1$ ones is equal to $2^{2n}$.
