Sum of Sequence of Odd Cubes

Theorem

 * $\displaystyle \sum_{j \mathop = 1}^n \left({2 j - 1}\right)^3 = 1^3 + 3^3 + 5^3 + \cdots + \left({2 n − 1}\right)^3 = n^2 \left({2 n^2 − 1}\right)$

Proof
Proof by induction:

For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{j \mathop = 1}^n \left({2 j - 1}\right)^3 = n^2 \left({2 n^2 − 1}\right)$

Basis for the Induction
$P(1)$ is the case:

and $P(1)$ is seen to hold.

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:
 * $\displaystyle \sum_{j \mathop = 1}^k \left({2 j - 1}\right)^3 = k^2 \left({2 k^2 − 1}\right)$

Then we need to show:
 * $\displaystyle \sum_{j \mathop = 1}^{k + 1} \left({2 j - 1}\right)^3 = \left({k + 1}\right)^2 \left({2 \left({k + 1}\right)^2 − 1}\right)$

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:
 * $\displaystyle \forall n \in \Z_{\ge 1}: \sum_{j \mathop = 1}^n \left({2 j - 1}\right)^3 = n^2 \left({2 n^2 − 1}\right)$