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