Combinatorics identity sum of

Question

Prove that:

$$\sum^{n}_{k=0}\binom{k}{2n-k}2^k = 2^{2n}$$

By using only combinatorics identities.

Answer 1

From this result we have

$$2^n=\sum_{k=0}^n2^{-k}\binom{n+k}k=\frac1{2^n}\sum_{k=0}^n2^{n-k}\binom{n+k}k=\frac1{2^n}\sum_{k=0}^n2^k\binom{2n-k}{n-k}=\frac1{2^n}\sum_{k=0}^n2^k\binom{2n-k}n\;,$$

so

$$\sum_{k=0}^n2^k\binom{2n-k}n=2^{2n}\;,$$

which I suspect is the intended result.