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

(Created page with "== Problem 1 == Simplify <cmath>\left(\frac{1\cdot2\cdot+2\cdot4\cdot8+\cdots+n\cdot2n\cdot4n}{1\cdot3\cdot9+2\cdot6\cdot18+\cdots+n\cdot3n\cdot9n}\right)^{1/3}</cmath>. == S...")
 
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== Problem 1 ==
 
== Problem 1 ==
 
Simplify
 
Simplify
<cmath>\left(\frac{1\cdot2\cdot+2\cdot4\cdot8+\cdots+n\cdot2n\cdot4n}{1\cdot3\cdot9+2\cdot6\cdot18+\cdots+n\cdot3n\cdot9n}\right)^{1/3}</cmath>.
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<cmath>\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}</cmath>.
  
 
== Solution ==  
 
== Solution ==  
None yet!
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<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}}

Latest revision as of 19:03, 7 April 2024

Problem 1

Simplify \[\left(\frac{1\cdot2\cdot4+2\cdot4\cdot8+\cdots+n\cdot2n\cdot4n}{1\cdot3\cdot9+2\cdot6\cdot18+\cdots+n\cdot3n\cdot9n}\right)^{1/3}\].

Solution

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


$=\left(\frac{2^3\cdot1+2^3\cdot2^3+2^3\cdot3^3+2^3\cdot4^3+\cdots+2^3n^3}{3^3+3^3\cdot2^3+3^3\cdot3^3+3^3\cdot4^3+\cdots+3^3n^3}\right)^{1/3}$


$=\left[\frac{2^3\cancel{(1^3+2^3+3^3+\cdots+n^3)}}{3^3\cancel{(1^3+2^3+3^3+\cdots+n^3)}}\right]^{1/3}$


$=\boxed{\frac{2}{3}}$


1975 Canadian MO (Problems)
Preceded by
First question
1 2 3 4 5 6 7 8 Followed by
Problem 2