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I'm trying to prove the statement $$ \sum_{\ell\text{ odd}}^{n} \binom{n}{\ell} = 2^{n-1} $$ by induction. The base case is $$\sum_{\ell\text{ odd}}^{1} \binom{n}{\ell} = \binom{1}{1} = 2^{1-1} = 1.$$

For the induction step, I'm assuming that $$\sum_{\ell\text{ odd}}^{n} \binom{n}{\ell} = 2^{n-1}$$ and trying to prove $$\sum_{\ell\text{ odd}}^{n+1} \binom{n+1}{\ell} = 2^{n+1-1} = 2^{n}$$

Because $$ \binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}$$

$$\sum_{\ell\text{ odd}}^{n+1} \binom{n+1}{\ell} = \sum_{\ell\text{ odd}}^{n+1} \binom{n}{\ell} + \sum_{\ell\text{ odd}}^{n+1} \binom{n}{\ell-1}.$$

The first part of the expression is equal to $$2^{n-1}$$ so I'm just trying to prove that $$\sum_{\ell\text{ odd}}^{n+1} \binom{n}{\ell-1} = \binom{n}{0} + \binom{n}{2} + \dotsb = 2^{n-1}$$ but I'm stuck at this step.

Xander Henderson
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rachel8841
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    https://math.stackexchange.com/questions/248245/exactly-half-of-the-elements-of-mathcalpa-are-odd-sized – user1001001 Jul 06 '20 at 15:50
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    As for completing your proof... recall that $\sum\limits_{l=0}^n \binom{n}{l} = 2^n$ where the index is not restricted to only odd values. – JMoravitz Jul 06 '20 at 15:53
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    $(1+1)^n=?$ and $(1-1)^n=?$ – Teddy38 Jul 06 '20 at 15:58
  • @JMoravitz I'm not sure how I would apply that question to this proof – rachel8841 Jul 06 '20 at 16:28
  • $\sum\limits_{l~\text{odd}}^{n+1}\binom{n}{l}+\sum\limits_{l~\text{odd}}^{n+1}\binom{n}{l-1} = \color{red}{\left(\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\dots\right)} + \color{blue}{\left(\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\dots\right)} = \color{blue}{\binom{n}{0}}+\color{red}{\binom{n}{1}}+\color{blue}{\binom{n}{2}}+\color{red}{\binom{n}{3}}+\dots = \sum\limits_{l=0}^n\binom{n}{l}=2^n$ This is just rearranging the terms in a finite sum, which is perfectly allowed. – JMoravitz Jul 06 '20 at 17:10
  • @JMoravitz I can prove $$\sum_{l=0}^{n} \binom{n}{l} = 2^{n}$$ but I don't know how to show that either the evens or the odds sum up to $2^{n-1}$ – rachel8841 Jul 06 '20 at 17:25
  • Your work you have already done along with the final line I just gave you directly imply that... of course it is a bit of a silly proof by induction since the induction hypothesis is never used in the proof... but your work showed that $\sum\limits_{l~\text{odd}}^{n+1}\binom{n+1}{l} = \sum\limits_{l~\text{odd}}^{n+1}\binom{n}{l}+\sum\limits_{l~\text{odd}}^{n+1}\binom{n}{l-1} = \sum\limits_{l=0}^n \binom{n}{l} = 2^n$ which completes the proof – JMoravitz Jul 06 '20 at 17:57
  • That said, I encourage you to look at the answers in the linked question for much more straightforward approaches rather than using Paschal's Identity as you had started. – JMoravitz Jul 06 '20 at 18:00

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