# Sum of Even Index Binomial Coefficients

## Theorem

Let $n > 0$ be a (strictly) positive integer.

Then:

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

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

That is:

$\dbinom n 0 + \dbinom n 2 + \dbinom n 4 + \dotsb = 2^{n - 1}$

## Proof 1

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:

$\dbinom n 0 + \dbinom n 1 + \dbinom n 2 + \dbinom n 3 + \dbinom n 4 + \cdots = 2^n$

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:

$\dbinom n 0 - \dbinom n 1 + \dbinom n 2 - \dbinom n 3 + \dbinom n 4 - \cdots = 0$

$2 \dbinom n 0 + 2 \dbinom n 2 + 2 \dbinom n 4 + \cdots = 2^n$

as the odd index coefficients cancel out.

Dividing by $2$ throughout gives us the result.

$\blacksquare$

## Proof 2

Let $N^*_n$ be the initial segment of natural numbers, one-based.

Let:

$E_n = \set {X : \paren {X \subset N^*_n} \land \paren {2 \divides \size X} }$
$O_n = \set {X : \paren {X \subset N^*_n} \land \paren {2 \nmid \size X} }$

That is:

$E_n$ is the set of all subsets of $N^*_n$ with an even number of elements
$O_n$ is the set of all subsets of $N^*_n$ with an odd number of elements.
$\ds \sum_{i \mathop \ge 0} \binom n {2 i} = 2^{n - 1} \iff \size {E_n} = 2^{n - 1}$

which is to be proved by induction below.

### Basis of the Induction

Let $n = 1$.

Then

$\size {E_n} = \size {\set \O} = 1$

and:

$2^{n - 1} = 2^{1 - 1} = 2^0 = 1$

This is the basis for the induction.

### Induction Hypothesis

This is the induction hypothesis:

$\size {E_k} = 2^{k - 1}$

So:

$\size {O_k} = \size {2^{N^*_k} \divides E_k} = 2^k - 2^{k-1} = 2^{k - 1}$

### Induction Step

This is the induction step:

 $\ds \size {E_{k + 1} }$ $=$ $\ds \size {\set {X: \paren {X \subset N^*_{k + 1} } \land \paren {2 \divides \size X} } }$ Definition of $E_n$ $\ds$ $=$ $\ds \size {\set {X: X \in E_k} \cup \set {X \cup \set {k + 1}: X \in O_k} }$ Construction of the set with smaller sets, considering the presence of the element $k+1$ $\ds$ $=$ $\ds \size {E_k} + \size {O_k}$ $E_k$ and $O_k$ are disjoint by definition $\ds$ $=$ $\ds 2^{k - 1} + 2^{k - 1}$ Induction Hypothesis $\ds$ $=$ $\ds 2^k$

The result follows by induction.

$\blacksquare$