Difference between revisions of "User talk:JBL"
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For every choice of <math>\omega</math>, <math>\overline \omega</math> is also a 10th root of unity, so if we take 2 copies of <math>X</math> we can pair up terms to get <math>2X = \sum_{\omega^{10} = 1} \frac 1{1 - \omega/2} + \frac1{1 - \overline{\omega}/2} = \sum_{\omega^{10} = 1} \frac {(1 - \omega/2) + (1 - \overline{\omega}/2)}{(1 - \omega/2)(1 - \overline{\omega}/2)}</math> | For every choice of <math>\omega</math>, <math>\overline \omega</math> is also a 10th root of unity, so if we take 2 copies of <math>X</math> we can pair up terms to get <math>2X = \sum_{\omega^{10} = 1} \frac 1{1 - \omega/2} + \frac1{1 - \overline{\omega}/2} = \sum_{\omega^{10} = 1} \frac {(1 - \omega/2) + (1 - \overline{\omega}/2)}{(1 - \omega/2)(1 - \overline{\omega}/2)}</math> | ||
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| + | == About CMO == | ||
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| + | I am sorry about CMO. What do you suggest I should do? Redirect? Recreate the pages? By the way "[[Cyprus Mathematical Olympiad]]" it is th official translation. The papers are in both English and Greek. | ||
Revision as of 16:10, 5 November 2006
For later use (Mock AIME 2006)
Let the ratio of consecutive terms of the sequence be 
. Then we have by the given that 
so 
and 
, where 
can be any of the tenth roots of unity.
Then the sum 
has value 
. Different choices of clearly lead to different values for 
, so we don't need to worry about the distinctness condition in the problem. Thus our final answer will be 
.
For every choice of , 
is also a 10th root of unity, so if we take 2 copies of 
we can pair up terms to get 
About CMO
I am sorry about CMO. What do you suggest I should do? Redirect? Recreate the pages? By the way "Cyprus Mathematical Olympiad" it is th official translation. The papers are in both English and Greek.
