I am confused on how to prove this combinations equation :(

Question

Prove that
$\sum_{k=0}^n {n+k\choose k}\Big(\frac{1}{2}\Big)^k = 2^n.$

I tried doing the following:

${n\choose 0}+\frac{n+1\choose 1}{2}+\frac{n+2\choose 2}{4}+\dots+\frac{n+n\choose n}{2^n}=2^n$
But then I got stuck... Can someone please provide me a hint (maybe I am approaching it the wrong way?)

You've tagged this question [tag:combinatorial-proofs]. Are you specifically looking for a combinatorial proof, or would any proof do? — user759562, Mar 22 '20 at 03:13
This is closely related to How to prove $f(n)=\sum_{k=0}^n\binom{n+k}{k}\left(\frac{1}{2}\right)^k=2^n$ without using the induction method? — robjohn, Mar 22 '20 at 04:00
Does this answer your question? Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$ — Lê Thành Đạt, Mar 22 '20 at 12:16

score 0 · Accepted Answer · answered Mar 22 '20 at 03:23

The result is true for when $n=0.$ Assume the result for $n,$ and I will show the result for $n+1.$

\begin{equation} \begin{split} \sum_{k=0}^{n+1} {n+k+1\choose k}\Big(\frac{1}{2}\Big)^k &= 1+\sum_{k=1}^{n+1} {n+k+1\choose k}\Big(\frac{1}{2}\Big)^k\\ &=1+\sum_{k=1}^{n+1} {n+k\choose k}\Big(\frac{1}{2}\Big)^k+\sum_{k=1}^{n+1} {n+k\choose k-1}\Big(\frac{1}{2}\Big)^k\\ &=(2^k+{2n+1\choose n+1}\Big(\frac12\Big)^{n+1})+\frac12\sum_{k=0}^n{n+k+1\choose k}\Big(\frac{1}{2}\Big)^k\\ &=2^k+\frac12\sum_{k=0}^{n+1} {n+k+1\choose k}\Big(\frac{1}{2}\Big)^k,\\ \end{split} \end{equation}

so $\sum_{k=0}^{n+1}{n+k+1\choose k}\Big(\frac{1}{2}\Big)^k=2^{n+1}.$

I am confused on how to prove this combinations equation :(

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