Sum of Sequence of Odd Cubes

Theorem

 * $\ds \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + \paren {2 n − 1}^3 = n^2 \paren {2 n^2 − 1}$

Proof
Proof by induction:

For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
 * $\ds \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = n^2 \paren {2 n^2 − 1}$

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

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

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:
 * $\ds \sum_{j \mathop = 1}^k \paren {2 j - 1}^3 = k^2 \paren {2 k^2 − 1}$

Then we need to show:
 * $\ds \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}^3 = \paren {k + 1}^2 \paren {2 \paren {k + 1}^2 − 1}$

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:
 * $\ds \forall n \in \Z_{\ge 1}: \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = n^2 \paren {2 n^2 − 1}$