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.