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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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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)

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The green result can be used to prove the red one, yes.

Start with the fact that $\binom{m}{k} = \binom{m}{m - k}$. If this isn't obvious to you, you can derive it either by looking at the usual formula for the binomial, or by considering that choosing the members of a subset is equivalent to choosing elements excluded from the subset.

You can use that fact to see that $\sum_{k=0}^n \binom{2n + 1}{k} = \sum_{k=0}^n \binom{2n + 1}{2n + 1 - k}$. Considering the set of numbers $\{2n + 1 - k \mid k \in \{ 0,\dots,n \}\}$, you can see that this set is the same as $\{n + 1,\dots, 2n + 1\}$, so another way to write $\sum_{k=0}^n \binom{2n + 1}{2n + 1 - k}$ is $\sum_{k=n + 1}^{2n + 1} \binom{2n + 1}{k}$.

Then, of course, $\sum_{k=0}^n \binom{2n + 1}{k} + \sum_{k=n + 1}^{2n + 1} \binom{2n + 1}{k} = \sum_{k=0}^{2n + 1} \binom{2n + 1}{k}$, which by the green equation is equal to $2^{2n + 1}$. But from the previous paragraph, we know the summands on the left are equal to each other, so they must each be equal to half of the result, or $2^{2n}$. This proves the red equation. (You need a little more to go the other way; using this line of reasoning starting with the red equation would only prove that the green equation holds for odd $n$.)

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