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 $n^3$.

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 $

$=$

$\ds n^3 - n^2 + n^2$

$\ds $

$=$

$\ds n^3$

$\blacksquare$

This entry was named for Nicomachus of Gerasa.