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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:
1=1
1+8=9
1+8+27=36

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

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Solution No Subject | Comment 4 of 11 |
<H2>The sum of cubes </H2>

Here are two well known important sums:

<CENTER><IMG alt=" begin{eqnarray*} 1 + 2 + 3 + 4 + ldots + n & = & frac{n(n + 1)}{2} \ 1^3 + 2^3 + 3^3 + 4^3 + ldots + n^3 & = & left[frac{n(n + 1)}{2}right]^2 end{eqnarray*} " src="file:///E:/Documents%20and%20Settings/Shireesha/Desktop/Just%20One%20Folder/Perplexus/The%20sum%20of%20cubes_files/cubes_1.gif" align=absMiddle> </CENTER> <CENTER>From this it is evident that the set of the numbers ( 1,2,3,.......n) has the property that the sum of cubes is equal to square of their sum.....Hence the results</CENTER> <CENTER> </CENTER> <CENTER>1=1 ...i.e 1 cubed =1
</CENTER> <CENTER>1+8=9...i.e (1 cubed + 2 cubed) = (1+2) whole squared</CENTER> <CENTER> </CENTER> <CENTER>1+8+27=36.. i.e (1 cubed + 2 cubed + 3 cubed ) = ( 1+2+3 ) whole squared.</CENTER> <CENTER> </CENTER> <CENTER>and so on....</CENTER> <CENTER> </CENTER> <CENTER>So it implies that the sum of consecutive perfect cubes (starting with 1) is always a perfect square.</CENTER> <CENTER> </CENTER> <CENTER> </CENTER>
  Posted by Pemmadu Raghu Ramaiah on 2005-01-09 02:31:13

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