Odd Number Theorem

Theorem

 * $\displaystyle \sum_{j \mathop = 1}^n \left({2j - 1}\right) = n^2$

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:


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

Proof by induction:
For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle n^2 = \sum_{j \mathop = 1}^n \left({2j - 1}\right)$

Basis for the Induction
$P(1)$ 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 $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:


 * $\displaystyle k^2 = \sum_{j \mathop = 1}^k \left({2j - 1}\right)$

Then we need to show:


 * $\displaystyle \left({k+1}\right)^2 = \sum_{j \mathop = 1}^{k+1} \left({2j - 1}\right)$

Induction Step
This is our induction step:

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

Therefore:
 * $\displaystyle \forall n \in \N: n^2 = \sum_{j \mathop = 1}^n \left({2j - 1}\right)$

Proof of Corollary
Follows directly.

Comment
What this shows is that every square number is the sum of a series of consecutive odd integers:


 * $n^2 = 1 + 3 + 5 + \cdots + \left({2n-1}\right)$