Sum of Sequence of Odd Squares/Formulation 2

Theorem

 * $\ds \forall n \in \Z_{> 0}: \sum_{i \mathop = 1}^n \paren {2 i - 1}^2 = \frac {4 n^3 - n} 3$

Proof
The proof proceeds by induction.

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
 * $\ds \sum_{i \mathop = 1}^n \paren {2 i - 1}^2 = \frac {4 n^3 - n} 3$

Basis for the Induction
$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown 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 the induction hypothesis:
 * $\ds \sum_{i \mathop = 1}^k \paren {2 i - 1}^2 = \frac {4 k^3 - k} 3$

from which it is to be shown that:
 * $\ds \sum_{i \mathop = 1}^{k + 1} \paren {2 i - 1}^2 = \frac {4 \paren {k + 1}^3 - \paren {k + 1} } 3$

Induction Step
This is the induction step:

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

Therefore:
 * $\ds \forall n \in \Z_{> 0}: \sum_{i \mathop = 1}^n \paren {2 i - 1}^2 = \frac {4 n^3 - n} 3$