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Sum of Cubes (Posted on 2004-05-25) Difficulty: 3 of 5
Prove that the sum of consecutive perfect cubes (starting with 1) is always a perfect square.

For example:

See The Solution Submitted by Gamer    
Rating: 3.4000 (5 votes)

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an arbitrary n solution... | Comment 6 of 11 |
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
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