Sum of Sequence of Cubes/Proof by Nicomachus

Theorem

$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$

$\forall n \in \N_{>0}: n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 + n - 1}$

Also by Nicomachus's Theorem, we have that the first term for $\paren {n + 1}^3$ is $2$ greater than the last term for $n^3$.

So if we add them all up together, we get:

$\ds \sum_{i \mathop = 1}^n i^3$

$=$

$\ds \sum_{\stackrel {1 \mathop \le i \mathop \le n^2 \mathop + n \mathop - 1} {i \text { odd} } } i$

... the sum of all the odd numbers up to $n^2 + n - 1$

$\ds $

$=$

$\ds \sum_{i \mathop = 1}^{\frac {n^2 \mathop + n} 2} 2 i - 1$

... that is, the first $\dfrac {n^2 + n} 2$ odd numbers

$\ds $

$=$

$\ds \paren {\frac {n^2 + n} 2}^2$

Hence the result.

$\blacksquare$