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

[n*(n+1)]^2 - [(n-1)*n]^2 = (n^2)*[(n+1)^2 - (n-1)^2] = 4*n^3. Hence

1+ 8+27+64+...+N^3=

={1 - 0} + {9 - 1} +{36 -9} +{100 - 36}+...+

+{[N*(N+1)/2]^2 - [(N-1)*N/2]^2} = [N*(N+1)/2]^2.

 The identity used follows from the useful "quarter square multiplier" formula (a+b)^2 - (a-b)^2 = 4*a*b.

Edited on May 26, 2004, 1:45 am

Posted by Richard on 2004-05-25 16:06:56