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

1=1                                           1²
1+8=9                                       3²
1+8+27=36                                6²
1+8+27+64=100                       10²
1+8+27+64+125=225               15²
1+8+27+64+125+216=441        21²

for ex. 6³ + (tr(5))² = (tr(6))²

given :tr(x) = (x²+x)/2
           tr(x-1) = (x²-x)/2

prove: n³+(tr(n-1))² = (tr(n))²

n³ + ((n²-n)/2)² = (n²+n)/2)²
n³ + (.5n² - .5n)² = (.5n² + .5n)²
n³ + .25n^4 - .5n³ + .25n² = .25n^4 + .5n³ + .25n²
n³ - .5n³ + .25n² = .5n³ + .25n²
n³ - .5n³ = .5n³
n³ = n³

Edited on May 25, 2004, 4:48 pm

Posted by Danny on 2004-05-25 16:45:17