# Odd Number Theorem

## Theorem

$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1} = n^2$

That is, the sum of the first $n$ odd numbers is the $n$th square number.

### Corollary

A recurrence relation for the square numbers is:

$S_n = S_{n - 1} + 2 n - 1$

## Visual Demonstration $\blacksquare$

## Proof

Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

$\ds n^2 = \sum_{j \mathop = 1}^n \paren {2 j - 1}$

### Basis for the Induction

$\map P 1$ is true, as this just says $1^2 = 1$.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$\ds k^2 = \sum_{j \mathop = 1}^k \paren {2 j - 1}$

Then we need to show:

$\ds \paren {k + 1}^2 = \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}$

### Induction Step

This is our induction step:

 $\ds \paren {k + 1}^2$ $=$ $\ds k^2 + 2 k + 1$ $\ds$ $=$ $\ds \sum_{j \mathop = 1}^k \paren {2 j - 1} + 2 k + 1$ Induction Hypothesis $\ds$ $=$ $\ds \sum_{j \mathop = 1}^k \paren {2 j - 1} + 2 \paren {k + 1} - 1$ $\ds$ $=$ $\ds \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\ds \forall n \in \N: n^2 = \sum_{j \mathop = 1}^n \paren {2 j - 1}$

$\blacksquare$

## Also presented as

This result can also be seen as:

$\ds \sum_{j \mathop = 0}^{n - 1} \paren {2 j + 1} = n^2$

## Historical Note

Francesco Maurolico's proof of the Odd Number Theorem appears to be the first proof by induction.

It appears in his Arithmeticorum Libri Duo, written in $1557$ but not published till $1575$.