Odd Number Theorem

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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

OddNumberTheorem.png

$\blacksquare$


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 $1575$.


Sources

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