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
<H2>The sum of cubes </H2>
Here are two well known important sums:
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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*}
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<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>
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<CENTER>1=1 ...i.e 1 cubed =1
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<CENTER>1+8=9...i.e (1 cubed + 2 cubed) = (1+2) whole squared</CENTER>
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<CENTER>1+8+27=36.. i.e (1 cubed + 2 cubed + 3 cubed ) = ( 1+2+3 ) whole squared.</CENTER>
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<CENTER>and so on....</CENTER>
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<CENTER>So it implies that the sum of consecutive perfect cubes (starting with 1) is always a perfect square.</CENTER>
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