Sum of Sequence of Odd Cubes

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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:

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

$\blacksquare$