Sum of Sequence of Cubes/Proof by Nicomachus

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Theorem

$\displaystyle \sum_{i=1}^n i^3 = \left({\sum_{i=1}^n i}\right)^2 = \frac{n^2 \left({n + 1}\right)^2} 4$

$\forall n \in \N^*: n^3 = \left({n^2 - n + 1}\right) + \left({n^2 - n + 3}\right) + \ldots + \left({n^2 + n - 1}\right)$

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

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

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \sum_{i=1}^n i^3$	$=$	$\displaystyle \sum_{\stackrel {1 \le i \le n^2 + n - 1} {i \text { odd} } } i$	$\displaystyle $	$\displaystyle $	$\displaystyle $	... the sum of all the odd numbers up to $n^2 + n - 1$
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \sum_{1=1}^{\frac {n^2 + n} 2} 2 i - 1$	$\displaystyle $	$\displaystyle $	$\displaystyle $	... that is, the first $\dfrac {n^2 + n} 2$ odd numbers
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \left({\frac {n^2 + n} 2}\right)^2$	$\displaystyle $	$\displaystyle $	$\displaystyle $	by the Odd Number Theorem

Hence the result.

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{i=1}^n i^3\)	\(=\)	\(\displaystyle \sum_{\stackrel {1 \le i \le n^2 + n - 1} {i \text { odd} } } i\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	... the sum of all the odd numbers up to $n^2 + n - 1$
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{1=1}^{\frac {n^2 + n} 2} 2 i - 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	... that is, the first $\dfrac {n^2 + n} 2$ odd numbers
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\frac {n^2 + n} 2}\right)^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the Odd Number Theorem