# 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.

Title | Nicomachus’ theorem |
---|---|

Canonical name | NicomachusTheorem |

Date of creation | 2013-03-22 18:07:12 |

Last modified on | 2013-03-22 18:07:12 |

Owner | PrimeFan (13766) |

Last modified by | PrimeFan (13766) |

Numerical id | 7 |

Author | PrimeFan (13766) |

Entry type | Theorem |

Classification | msc 11A25 |

Related topic | CubeOfAnInteger |