Combinatorics Binomial How to prove it?

Question

$$ \sum_{k=n}^{2n} \binom{k}{n} 2^{-k}=1 $$ Anyone can help me? How to prove it?

Answer 1

Under the convention that $\binom{k}{n}=0$ if $k<n$ we can deduce:

$\begin{aligned}\sum_{k\leq2n}\binom{k}{n}2^{-k} & =\sum_{k\leq2n}\left[\binom{k-1}{n-1}+\binom{k-1}{n}\right]2^{-k}\\ & =\sum_{k\leq2n}\binom{k-1}{n-1}2^{-k}+\sum_{k\leq2n}\binom{k-1}{n}2^{-k}\\ & =\sum_{k\leq2n-1}\binom{k}{n-1}2^{-k-1}+\sum_{k\leq2n-1}\binom{k}{n}2^{-k-1}\\ & =\sum_{k\leq2n-2}\binom{k}{n-1}2^{-k-1}+\binom{2n-1}{n-1}2^{-2n}+\sum_{k\leq2n}\binom{k}{n}2^{-k-1}-\binom{2n}{n}2^{-2n-1}\\ & =\frac{1}{2}\sum_{k\leq2n-2}\binom{k}{n-1}2^{-k}+\frac{1}{2}\sum_{k\leq2n}\binom{k}{n}2^{-k} \end{aligned} $

The last equality implies that: $$\sum_{k\leq2n}\binom{k}{n}2^{-k}=\sum_{k\leq2n-2}\binom{k}{n-1}2^{-k}$$ so now induction can be applied to prove that $$\sum_{k\leq2n}\binom{k}{n}2^{-k}=1$$