# 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:

 $\displaystyle \sum_{j \mathop = 1}^1 \left({2 j - 1}\right)^3$ $=$ $\displaystyle 1^3$ $\displaystyle$ $=$ $\displaystyle 1^2 \left({2 \times 1^2 - 1}\right)$

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:

 $\displaystyle \sum_{j \mathop = 1}^{k + 1} \left({2 j - 1}\right)^3$ $=$ $\displaystyle \sum_{j \mathop = 1}^k \left({2 j - 1}\right)^3 + \left({2 k + 1}\right)^3$ $\displaystyle$ $=$ $\displaystyle k^2 \left({2 k^2 − 1}\right)+ \left({2 k + 1}\right)^3$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle 2 k^4 + 8 k^3 + 11 k^2 + 6 k + 1$ multiplying out $\displaystyle$ $=$ $\displaystyle \left({k + 1}\right)^2 \left({2 k^2 + 4 k + 1}\right)$ extracting $\left({k + 1}\right)^2$ as a factor $\displaystyle$ $=$ $\displaystyle \left({k + 1}\right)^2 \left({2 \left({k + 1}\right)^2 - 1}\right)$

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

Therefore:

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

$\blacksquare$