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$\sum\limits_{l\text{ odd}}\dbinom{n}{l} = \dbinom{n}{l} + \dbinom{n}{3} + ... = 2^{n-1}$

This has to be done via induction, but i'm not sure how. The base case is easily proved by taking n = 1. The induction argument, however, requires the $\sum\limits_{l\text{ odd}}\dbinom{n}{l} = 2^{n-1}$ to become $\sum\limits_{l\text{ odd}}\dbinom{n+1}{l} = 2^n$. I'm looking for only a slight hint as to how this may be done.

robjohn
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Aayush Agrawal
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  • "This has to be done via induction" <--- why? – Martin Brandenburg Jan 17 '16 at 09:15
  • Related: http://math.stackexchange.com/questions/1615343/the-equation-to-be-proved-is-sum-j-0n-1j-dbinomnj-dbinomn0 – Batominovski Jan 17 '16 at 09:20
  • @MartinBrandenburg The problem is in a induction part of the book, and all the other questions have used induction. So it's safe to assume that the author wanted to use induction. – Aayush Agrawal Jan 17 '16 at 09:22
  • And your accepted answer doesn't involve induction. – Martin Brandenburg Jan 17 '16 at 10:05
  • That's true. I was confused about which answer to select, but in the end i asked for a hint for how to solve it, and Rob gave that hint. Then i asked him to elaborate and since there was only one step left, he had to give away the main proof itself. I thought that it should quality to be the correct answer because of that. – Aayush Agrawal Jan 17 '16 at 10:20

2 Answers2

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Hint: Try expanding $(1+1)^n$ and $(1-1)^n$ using the Binomial Theorem.

More after comments $$ \begin{align} 2^n &=(1+1)^n\\ &=\sum_{k=0}^n\binom{n}{k}\tag{1} \end{align} $$ $$ \begin{align} 0^n &=(1-1)^n\\ &=\sum_{k=0}^n\binom{n}{k}(-1)^k\tag{2} \end{align} $$ Subtract $(2)$ from $(1)$ and divide by $2$: $$ 2^{n-1}=\sum_{k=0}^n\binom{n}{k}\frac{1-(-1)^k}2\tag{3} $$ Consider what $\frac{1-(-1)^k}2$ is for even and odd $k$.

robjohn
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One way: $\binom{n+1}{l} = \binom{n}{l} + \binom{n}{l-1}$, and note that if we assume $\sum_{l \text{ odd}} \binom{n}{l} = 2^{n-1}$ we also have $\sum_{l \text{ even}} \binom{n}{l} = 2^{n-1}$, as $\sum_{l} \binom{n}{l} = 2^{n}$.

This will allow you to use an induction proof.

Henno Brandsma
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