Sum of Odd Index Binomial Coefficients

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \sum_{i \mathop \ge 0} \binom n {2 i + 1} = 2^{n - 1}$

where $\dbinom n i$ is a binomial coefficient.


That is:

$\dbinom n 1 + \dbinom n 3 + \dbinom n 5 + \dotsb = 2^{n - 1}$


Proof

From Sum of Binomial Coefficients over Lower Index we have:

$\ds \sum_{i \mathop \in \Z} \binom n i = 2^n$

That is:

$\dbinom n 0 + \dbinom n 1 + \dbinom n 2 + \dbinom n 3 + \cdots + \dbinom n n = 2^n$

as $\dbinom n i = 0$ for $i < 0$ and $i > n$.

This can be written more conveniently as:

$(1): \quad \dbinom n 0 + \dbinom n 1 + \dbinom n 2 + \dbinom n 3 + \dbinom n 4 + \cdots = 2^n$


Similarly, from Alternating Sum and Difference of Binomial Coefficients for Given n we have:

$\ds \sum_{i \mathop \in \Z} \paren {-1}^i \binom n i = 0$

That is:

$(2): \quad \dbinom n 0 - \dbinom n 1 + \dbinom n 2 - \dbinom n 3 + \dbinom n 4 - \cdots = 0$

Subtracting $(2)$ from $(1)$, we get:

$2 \dbinom n 1 + 2 \dbinom n 3 + 2 \dbinom n 5 + \cdots = 2^n$

as the even index coefficients cancel out.

Dividing by $2$ throughout gives us the result.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Sum_of_Odd_Index_Binomial_Coefficients&oldid=596898"