TITLE: How to prove that $\sum_{i=0}^n 2^i\binom{2n-i}{n} = 4^n$.
QUESTION [15 upvotes]: So I've been struggling with this sum for some time and I just can't figure it out. I tried proving by induction that if the sum above is a $S_n$ then $S_{n+1} = 4S_n$, but I didn't really succeed so here I am. Thanks in advance.

REPLY [4 votes]: Here is a way to rephrase Zac's proof as a combinatorial argument.

Question: How many binary sequences of length $2n+1$ have more ones than zeroes?

Answer $1$: Half of the sequences have more ones than zeroes, since a sequence has more ones than zeroes if and only if its complement does not. There are $2^{2n+1}$ sequences total; half of this is $2^{2n}$.
Answer $2$: Such a sequence will have at least $n+1$ ones. Let us count how many such sequences there are such that the $(n+1)^{st}$ instance of one (reading from left to right) occurs at spot number $2n+1-k$. Among the first $2n-k$ symbols, there will be exactly $n$ ones, whose locations can be chosen in $\binom{2n-k}n$ ways. Symbol number $2n-k+1$ must be a one, and then the remaining $k$ symbols can be chosen arbitrarily in $2^k$ ways. Summing over $k$, the number of possible sequences is $\sum_{k=0}^n \binom{2n-k}{n}2^k$. 
Note the bounds on $k$; $k=0$ corresponds to the situation where the $(n+1)^{st}$ one occurs at spot $2n+1$, and $k=n$ corresponds to the case where the $(n+1)^{st}$ one occurs at spot $n+1$. These are indeed the furthest possible locations.