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.

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:

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

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

Also presented as
This result can also be seen as:
 * $\ds \sum_{j \mathop = 0}^{n - 1} \paren {2 j + 1} = n^2$