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

Revision as of 12:43, 24 January 2018

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

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

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

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

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

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

@@ Line 4: / Line 4: @@
 == Solution ==
-None yet!
+<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>
+<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>
+<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>
+<math>\boxed{\frac{2}{3}}</math>
 {{Old CanadaMO box|before=First question|num-a=2|year=1975}}