# 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:

 $\ds \sum_{i \mathop = 1}^1 \paren {2 i - 1}^2$ $=$ $\ds \paren {2 \times 1 - 1}^2$ $\ds$ $=$ $\ds 1^2$ $\ds$ $=$ $\ds \dfrac {\paren {4 \times 1 - 1}^3} 3$

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:

 $\ds \sum_{i \mathop = 1}^{k + 1} \paren {2 i - 1}^2$ $=$ $\ds \sum_{i \mathop = 1}^k \paren {2 i - 1}^2 + \paren {2 \paren {k + 1} - 1}^2$ $\ds$ $=$ $\ds \frac {4 k^3 - k} 3 + \paren {2 k + 1}^2$ Induction Hypothesis $\ds$ $=$ $\ds \frac {4 k^3 - k + 12 k^2 + 12 k + 3} 3$ $\ds$ $=$ $\ds \frac {\paren {k + 1} \paren {4 k^2 + 8 k + 3} } 3$ $\ds$ $=$ $\ds \frac {\paren {k + 1} \paren {4 \paren {k + 1}^2 - 1} } 3$ $\ds$ $=$ $\ds \frac {4 \paren {k + 1}^3 - \paren {k + 1} } 3$

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

Therefore:

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

$\blacksquare$