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# Nicomachus’ theorem

Theorem (Nicomachus). The sum of the cubes of the first $n$ integers is equal to the square of the $n$th triangular number. To put it algebraically,

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

###### Proof.

There are several formulas for the triangular numbers. Gauss figured out that to compute

$\sum_{{i=1}}^{n}i,$ |

one can, instead of summing the numbers one by one, pair up the numbers thus: $1+n$, $2+(n-1)$, $3+(n-2)$, etc., and each of these sums has the same result, namely, $n+1$. Since there are $n$ of these sums, carrying this all the way through to the end, we are in effect squaring $n+1$, which is $(n+1)^{2}=(n+1)(n+1)=n^{2}+n$. But this is redundant, since it includes both $1+n$ and $n+1$, both $2+(n-1)$ and $(n-1)+2$, etc., in effect, each of these twice. Therefore,

$\frac{n^{2}+n}{2}=\sum_{{i=1}}^{n}i.$ |

As Sir Charles Wheatstone proved, we can rewrite $i^{3}$ as

$\sum_{{j=1}}^{i}(2ij+i).$ |

That sum can always be rewritten as a sum of odd terms, namely

$\sum_{{k=1}}^{i}(i^{2}+i+2k).$ |

Thus, the sum of the first $n$ cubes is in fact also

$\sum_{{i=0}}^{{n-1}}(2i+1).$ |

The sum of the first $n-1$ odd numbers is $n^{2}$, and therefore

$\sum_{{i=1}}^{n}i^{3}=\left(\sum_{{i=1}}^{n}i\right)^{2},$ |

as the theorem states. ∎

For example, the sum of the first four cubes is 1 + 9 + 27 + 64 = 100. This is also equal to 1 + 3 + 5 + 7 + 9 + 11 + 13 + 17 + 19 = 100. The square root of 100 is 10, the fourth triangular number, and indeed 10 = 1 + 2 + 3 + 4.

## Mathematics Subject Classification

11A25*no label found*

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