# Nicomachus's Theorem/Proof 2

## Theorem

 $\ds 1^3$ $=$ $\ds 1$ $\ds 2^3$ $=$ $\ds 3 + 5$ $\ds 3^3$ $=$ $\ds 7 + 9 + 11$ $\ds 4^3$ $=$ $\ds 13 + 15 + 17 + 19$ $\ds \vdots$  $\ds$

In general:

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

In particular, the first term for $\paren {n + 1}^3$ is $2$ greater than the last term for $n^3$.

## Proof

From the definition:

$\paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 + n - 1}$

can be written:

$\paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 - n + 2 n - 1}$

Writing this in sum notation:

 $\ds$  $\ds \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 - n + 2 n - 1}$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n \paren {n^2 - n + 2 k - 1}$ $\ds$ $=$ $\ds n \paren {n^2 - n} + \sum_{k \mathop = 1}^n \paren {2 k - 1}$ $\ds$ $=$ $\ds n^3 - n^2 + n^2$ Odd Number Theorem $\ds$ $=$ $\ds n^3$

$\blacksquare$

## Source of Name

This entry was named for Nicomachus of Gerasa.