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Prove using combinatorics $\sum\limits_{r=0}^n2^{n-r}*\binom{n+r}{n}=2^{2n}$

The right side is choosing some persons from $2n$ people but I can't find a way to explain the left side it has both combination and power of $2$.Any hints?

Taha Akbari
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  • After a change of variables $r\mapsto n-r$ your problem is solved here: http://math.stackexchange.com/questions/508944/how-prove-this-sum-k-0n-frac-binom2n-kn22n-k-1 –  Apr 05 '17 at 13:23
  • If anyone needs detail on user940's comment, see https://math.codidact.com/posts/282644. –  Jul 11 '21 at 19:33

2 Answers2

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Consider the set $S$ of all possible bitstrings of length $2n$. Given a bitstring $s \in S$, define $k$-prefix of $s$ as the shortest prefix bitstring of $s$ with either exactly $k$ $0$s or exactly $k$ $1$s. Since we are considering the shortest prefix, a $k$-prefix with exactly $k$ 1s ends with bit $1$; a $k$-prefix with exactly $k$ 0s ends with bit $0$.

Now, consider $(n+1)$-prefixes. Any $(n+1)$-prefix must be of length at least $n+1$ and at most $2n$. For a given length $n+1+r$ of $(n+1)$-prefixes ($0 \le r \le n-1$), there are $2 \cdot {{n+r}\choose{n}}$ possible such $(n+1)$-prefixes. Each $s \in S$ has a unique $(n+1)$-prefix, unless $s$ has exactly $n$ 0s and $n$ 1s. There are $2^{n-r-1}$ bitstrings in $S$ that share a given $(n+1)$-prefix of length $n+1+r$. The total number of bitstrings in $S$ is therefore: $$ {{2n}\choose{n}} + \sum_{r=0}^{n-1}2{{n+r}\choose{n}}\cdot 2^{n-r-1} = \sum_{r=0}^{n}{{n+r}\choose{n}}\cdot 2^{n-r} $$

wto
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  • May I ask: a prefix starts an the beginning of the string $s\in S,$ and you're choosing just the first $n,0$'s or $1'$s by $\binom{n+r}n$ to count all the $(n+1)$ prefixes of the length $n+1+r$? And $\binom{2n}n$ is the number of bit strings without a unique $k$ prefix? – PinkyWay Oct 07 '21 at 15:44
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Hint:

Consider monotonic lattice paths from $(0,0)$ to the line $y=2n-x$ that pass through the line $\smash{x=n+\frac{1}{2}}$ and that pass through the line $\smash{y=n+\frac{1}{2}}$ and those directly to the point $(n,n)$.

N. Shales
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