Prove that the sum of consecutive perfect cubes (starting with 1) is always a perfect square.

For example:
1=1
1+8=9
1+8+27=36

The sum of the first n cubes is the square of the n-th triangular number: [n(n+1)/2]^2.

The formula is obviously correct for n=1. Assuming it's correct for all values up to n-1 we can show it's correct for n.

If the sum of all cubes up to n-1 is [(n-1)n/2]^2 adding n^3, the result can be rearranged to give [n(n+1)/2]^2 as desired.

Posted by e.g. on 2004-05-25 13:10:38