Sum of Consecutive Cubes

Date: 05/11/2000 at 20:57:35
From: cuong
Subject: Sum of consecutive cube number

Prove that the sum of consecutive cubes equals a square. That is:

     1^3 + 2^3 + 3^3 + ... + n^3 = m^2

Here is what I did:

Let n = 1:

     1^3 = 1 = 1^2 (true)

Let n = 2:

     1^3 + 2^3 = 9 = 3^2 (true)

Assume it is true for n = k:

     1^3 + 2^3 + ... + k^3 = m^2   ..................[1]

Prove it is true for n = k+1:

     1^3 + 2^3 + ... + k^3 + (k+1)^3 = ?   ..........[2]

Replace [1] in [2]:

     m^2 + (k+1)^3 = (???)   ........................[3]

Consider the left-hand side of [3] for a new square. Answer by 
mathematical induction.

Can you give me an idea?

Date: 05/11/2000 at 22:14:48
From: Doctor Schwa
Subject: Re: Sum of consecutive cube number

The proof is a lot easier if you know the relation between m and k, so 
you might want to try a few examples to see if you can guess the 
relation, and then prove it by induction.

Or, if you want me to give it away, here it is ... stop reading now if 
you want to figure it out for yourself ... the sum of the first k 
cubes is the square of the number: k(k+1)/2.

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/