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Comments on $\sum_{k=0}^{n} \binom{n}{k}=2^{n} \overset{?}{\iff} \sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}$

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$\sum_{k=0}^{n} \binom{n}{k}=2^{n} \overset{?}{\iff} \sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}$

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Jack D'Aurizio narratively proved $\color{red}{\sum\limits_{k=0}^{n} \binom{2n+1}{k}=2^{2n}}$. Is this red equation related, and can it be transmogrified, to $\color{limegreen}{\sum\limits_{k=0}^{n} \binom{n}{k}=2^{n}}$?

I started my attempt by substituting $n = m/2$, because the RHS of the green target equation has the form $\color{limegreen}{2^?}$. Then $\color{red}{\sum\limits_{k=0}^{n} \binom{2n+1}{k}=2^{2n} \implies \sum\limits_{k=0}^{m/2} \binom{m+1}{k}=2^{m}}$. But now what do I do? I can't rewrite $m$, again because the RHS of the target equation has the form $\color{limegreen}{2^?}$.

  1. Give a story proof that $\sum\limits_{k=0}^{n} \binom{n}{k}=2^{n}$.

Blitzstein. Introduction to Probability (2019 2 ed). p 35.

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2 comment threads

Technically $\Leftrightarrow$ just means that the statements on either side have the same truth value... (1 comment)
If the quoted exercise is the true problem and this is an XY question, it's far simpler to consider t... (1 comment)
If the quoted exercise is the true problem and this is an XY question, it's far simpler to consider t...
Peter Taylor‭ wrote over 3 years ago

If the quoted exercise is the true problem and this is an XY question, it's far simpler to consider the basic combinatorial meaning of $\binom{n}{k}$.