# Difference between revisions of "1975 Canadian MO Problems/Problem 1"

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== Solution == | == Solution == | ||

− | + | <math>\left(\frac{1\cdot2\cdot4+2\cdot4\cdot8+\cdots+n\cdot2n\cdot4n}{1\cdot3\cdot9+2\cdot6\cdot18+\cdots+n\cdot3n\cdot9n}\right)^{1/3}</math> | |

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+ | <math>\left(\frac{2^3\cdot1+2^3\cdot2^3+2^3\cdot3^3+2^3\cdot4^3+\cdots+2^3n^3}{3^3+3^3\cdot2^3+3^3\cdot3^3+3^3\cdot4^3+\cdots+3^3n^3}\right)^{1/3}</math> | ||

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+ | <math>\left[\frac{2^3\cancel{(1^3+2^3+3^3+\cdots+n^3)}}{3^3\cancel{(1^3+2^3+3^3+\cdots+n^3)}}\right]^{1/3}</math> | ||

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+ | <math>\boxed{\frac{2}{3}}</math> | ||

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{{Old CanadaMO box|before=First question|num-a=2|year=1975}} | {{Old CanadaMO box|before=First question|num-a=2|year=1975}} |