Post History
#5: Post edited
by
DNB
·
2021-07-17T05:04:50Z (almost 3 years ago)
[Jack D'Aurizio narratively proved](https://math.stackexchange.com/a/2327533) $\color{red}{\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}}$. Is this red equation related, and can it be transmogrified, to $\color{limegreen}{\sum_{k=0}^{n} \binom{n}{k}=2^{n}}$?I started my attempt by substituting $n = m/2$, because the RHS of the target equation has the form $\color{limegreen}{2^?}$. Then $\color{red}{\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n} \implies \sum_{k=0}^{m/2} \binom{m+1}{k}=2^{m}}$. But now what do I do? I can't rewrite $m$, again because the RHS of the target equation has the form $\color{limegreen}{2^?}$.>15. Give a story proof that $\sum_{k=0}^{n} \binom{n}{k}=2^{n}$.- Blitzstein. *Introduction to Probability* (2019 2 ed). p 35.
- [Jack D'Aurizio narratively proved](https://math.stackexchange.com/a/2327533) $\color{red}{\sum\limits_{k=0}^{n} \binom{2n+1}{k}=2^{2n}}$. 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 $\color{limegreen}{2^?}$.
- >15. Give a story proof that $\sum\limits_{k=0}^{n} \binom{n}{k}=2^{n}$.
- Blitzstein. *Introduction to Probability* (2019 2 ed). p 35.
#4: Post edited
by
DNB
·
2021-07-11T06:02:24Z (almost 3 years ago)
[Here's a story proof](https://math.stackexchange.com/a/2327533) for $\color{red}{\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}}$. Is this red equation related, and can it be transmogrified, to $\color{limegreen}{\sum_{k=0}^{n} \binom{n}{k}=2^{n}}$?- I started my attempt by substituting $n = m/2$, because the RHS of the target equation has the form $\color{limegreen}{2^?}$. Then $\color{red}{\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n} \implies \sum_{k=0}^{m/2} \binom{m+1}{k}=2^{m}}$. But now what do I do? I can't rewrite $m$, again because the RHS of the target equation has the form $\color{limegreen}{2^?}$.
- >15. Give a story proof that $\sum_{k=0}^{n} \binom{n}{k}=2^{n}$.
- Blitzstein. *Introduction to Probability* (2019 2 ed). p 35.
- [Jack D'Aurizio narratively proved](https://math.stackexchange.com/a/2327533) $\color{red}{\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}}$. Is this red equation related, and can it be transmogrified, to $\color{limegreen}{\sum_{k=0}^{n} \binom{n}{k}=2^{n}}$?
- I started my attempt by substituting $n = m/2$, because the RHS of the target equation has the form $\color{limegreen}{2^?}$. Then $\color{red}{\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n} \implies \sum_{k=0}^{m/2} \binom{m+1}{k}=2^{m}}$. But now what do I do? I can't rewrite $m$, again because the RHS of the target equation has the form $\color{limegreen}{2^?}$.
- >15. Give a story proof that $\sum_{k=0}^{n} \binom{n}{k}=2^{n}$.
- Blitzstein. *Introduction to Probability* (2019 2 ed). p 35.
#3: Post edited
by
DNB
·
2021-07-11T05:51:59Z (almost 3 years ago)
$\sum_{k=0}^{n} \binom{n}{k}=2^{n} \iff^{?}\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}$
- $\sum_{k=0}^{n} \binom{n}{k}=2^{n} \overset{?}{\iff} \sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}$
#2: Post edited
by
DNB
·
2021-07-11T05:48:32Z (almost 3 years ago)
$\sum_{k=0}^{n} \binom{n}{k}=2^{n}$ vs. $\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}$
- $\sum_{k=0}^{n} \binom{n}{k}=2^{n} \iff^{?}\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}$
#1: Initial revision
by
DNB
·
2021-07-11T05:48:01Z (almost 3 years ago)
$\sum_{k=0}^{n} \binom{n}{k}=2^{n}$ vs. $\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}$
[Here's a story proof](https://math.stackexchange.com/a/2327533) for $\color{red}{\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}}$. Is this red equation related, and can it be transmogrified, to $\color{limegreen}{\sum_{k=0}^{n} \binom{n}{k}=2^{n}}$?
I started my attempt by substituting $n = m/2$, because the RHS of the target equation has the form $\color{limegreen}{2^?}$. Then $\color{red}{\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n} \implies \sum_{k=0}^{m/2} \binom{m+1}{k}=2^{m}}$. But now what do I do? I can't rewrite $m$, again because the RHS of the target equation has the form $\color{limegreen}{2^?}$.
>15. Give a story proof that $\sum_{k=0}^{n} \binom{n}{k}=2^{n}$.
Blitzstein. *Introduction to Probability* (2019 2 ed). p 35.