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

Latest revision as of 13:12, 17 June 2011

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

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

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 ==Solution==
-<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>
+<math> \frac{1^{3}+2^{3}+3^{3}+\cdots+x^{3}}{1+2+3+\cdots+x}=\frac{(1+2+3+\cdots+x)^2}{1+2+3+\cdots+x}=1+2+3+\cdots+x</math>
-<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>
+<math>1+2+3+\cdots+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>