I'm trying to prove the following: $$\sum_{k=0}^n 2^k\binom{2n-k}{n}=4^n$$
I've thought about induction, but there's not a very nice way to change the LHS from the $n$ case to the $n+1$ case by multiplying by $4$, at least, none that I can see.
I've also thought about trying to put a combinatorial argument together, by somehow arguing that the LHS counts $n$-letter words made from a $4$-letter alphabet, but I can't get the set being counted to partition in a way that sensibly represents the different terms of the sum.
I believe the identity is true, not only because I've tried a few small values of $n$, but also because Wolfram Alpha simplifies it for me. It just won't tell me how.
Thanks in advance for any insights.
Additional context: This identity arose while studying answers to this probability question: Expected number of draws to find a match