I am trying to show that $$\sum_{k=0}^n 2^{-k}{k+n \choose k}=2^n.$$
I try $n=0$: $$ \sum_{k=0}^0 2^{-k}{k+n \choose k}=2^0{0 \choose 0}=1=2^0, $$ and $n=1$: $$ \sum_{k=0}^1 2^{-k}{k+n \choose k}=1+ 2^{-1}{1+1 \choose 1}=1+\frac{1}{2}\cdot 2=1+1=2^1, $$ so far I have a solid understanding of how this is working. It seems like induction might be a good way to go about finishing this, but I can't quite see how to make the connection of how assuming $n$ is true implies $n+1$ is true. Any advice?
Or alternatively, there may be a simple combinatorial proof that does not rely on induction, and instead utilizes a clever rearrangement. (I have rather poor intuition of these types of proofs, so that's why I tend to prefer induction, where there are clear steps towards a solution.)
