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Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 17, 2011"

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==Solution==
 
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<math> \frac{1^{3}+2^{3}+3^{3}+...+x^{3}}{1+2+3+...+x}=\frac{(1+2+3+...+x)^2}{1+2+3+...+x}=1+2+3+...+x</math>
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<math>1+2+3+...+x=\frac{x\cdot(x+1)}{2}</math> Subbing in the value of <math>x</math> we get,<math>\frac{9001\cdot9002}{2}=\boxed{40513501}</math>

Revision as of 22:31, 16 June 2011

Problem

AoPSWiki:Problem of the Day/June 17, 2011

Solution

This Problem of the Day needs a solution. If you have a solution for it, please help us out by adding it.

$\frac{1^{3}+2^{3}+3^{3}+...+x^{3}}{1+2+3+...+x}=\frac{(1+2+3+...+x)^2}{1+2+3+...+x}=1+2+3+...+x$


$1+2+3+...+x=\frac{x\cdot(x+1)}{2}$ Subbing in the value of $x$ we get,$\frac{9001\cdot9002}{2}=\boxed{40513501}$

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