How prove $\sum_{k=0}^n {2n-k\choose n-k} 2^k=4^n$

Question

show that

$$\sum_{k=0}^n {2n-k\choose n-k} 2^k=4^n$$

Have you other good methods, such as calculus methods?

this problem is from How prove this $\sum_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1 $

Answer 1

It is enough to show that $\sum_{k=0}^{n}\left({n+k\atop k}\right)2^{-k}=2^{n}$. This because $\sum_{k=0}^{n}\left({2n-k\atop n-k}\right)2^{k}=2^{n}\sum_{k=0}^{n}\left({n+k\atop k}\right)2^{-k}$. (switch order of addition)

Go out from $\sum_{k=0}^{n}\left({n+k\atop k}\right)2^{-k}=2^{n}$ and set $s:=\sum_{k=0}^{n+1}\left({n+1+k\atop k}\right)2^{-k}$

Then $s=1+\sum_{k=1}^{n+1}\left({n+k\atop k-1}\right)2^{-k}+\sum_{k=1}^{n+1}\left({n+k\atop k}\right)2^{-k}=1+2^{-1}\sum_{k=0}^{n}\left({n+1+k\atop k}\right)2^{-k}+\sum_{k=1}^{n+1}\left({n+k\atop k}\right)2^{-k}=1+2^{-1}\left(s-\left({2n+2\atop n+1}\right)2^{-n-1}\right)+2^{n}-1+\left({2n+1\atop n+1}\right)2^{-n-1}=2^{n}+2^{-1}s$ leading to $s=2^{n+1}$

Note that $\left({2n+2\atop n+1}\right)=2\left({2n+1\atop n+1}\right)$