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

Let F(n) = 1^3 + 2^3 + ....+ n^3, say.

Then, from the given example:
F(1) = 1 = (1*2/2)^2
F(2) = 9 = (2*3/2)^2
 F(3) = 36 = (3*4/2)^2  

This leads us to conjecture that:

F(n) = (n(n+1)/2)^2 ....(#)

Now, for n=t, we have:

F(t) = (t(t+1)/2)^2

-> F(t+1) = (t+1)^3 + F(t)
= (t+1)^3 + (t(t+1)/2)^2
= (1/4)*(t+1)^2*(4t+4+ t^2)
= (1/4)*(t+1)^2*(t+2)^2
= ((t+1)(t+2)/2)^2

Thus, the result holds for n=t+1 iff it holds for n=t.

It has already been shown that the result holds for n=1,2,3.

Therefore, the result (#) holds for any positive integer value of n.

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

Posted by K Sengupta on 2008-11-27 13:24:04