combinatorial proof that $\sum_{i=0}^n {n+i\choose i}\frac{1}{2^i} = 2^n$

Question

Give a combinatorial proof that $\displaystyle\sum_{i=0}^n {n+i\choose i}\frac{1}{2^i} = 2^n$.

I'm not sure if Pascal's identity is useful here. Or perhaps there is a way involving binary strings? $2^n$ is the number of binary strings of length $n$, so if there was some way to decompose these strings into disjoint sets $B_i$ with cardinality ${n+i\choose i}\frac{1}{2^i}$, a proof using that method might work. ${n+i\choose i}$ is the number of binary strings of length $n+i$ with exactly $i$ ones, since one can choose $i$ of the $n+i$ positions to be ones in ${n+i\choose i}$ ways and make the rest zeroes in one way. Then we divide by the number of binary strings of length $i$, though I'm not sure how to deduce the combinatorial significance of this. I'm most likely thinking of this the wrong way.

Answer 1

This is equivalent to $\sum_{i=0}^n\binom{n+i}{i}2^{n-i}=4^n$. Both sides count the number of length-$(2n+1)$ bit strings with at least $n+1$ ones; take $n+i+1$ to be the number of bits required to get the $(n+1)$th $1$, then choose which $i$ of the first $n+i$ bits are $0$s.