# 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:

 $\ds \sum_{j \mathop = 1}^1 \paren {2 j - 1}^3$ $=$ $\ds 1^3$ $\ds$ $=$ $\ds 1^2 \paren {2 \times 1^2 - 1}$

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:

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

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

Therefore:

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

$\blacksquare$