Prove that the sum of consecutive perfect cubes (starting with 1) is always a perfect square.
The sum of the cubes from one to n is equal to (n^2*(n+1)^2)/4 (this can be shown through induction).
Consider the sum of integers from 1 to n, squared. This is equal to (n(n+1)/2)*(n(n+1)/2) through induction.
Thus, the sum of the cubes from 1 to n is equal to the sum from 1 to n, squared.
Posted by ryan
on 2005-08-14 21:55:39