Difference between revisions of "User talk:JBL"
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| + | == For later use (Mock AIME 2006) == | ||
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| + | Let the [[ratio]] of consecutive terms of the sequence be <math>r \in \mathbb{C}</math>. Then we have by the given that <math>1 = a_{10} = r^{10} a_0 = 1024r^{10}</math> so <math>r^{10} = 2^{-10}</math> and <math>r = \frac \omega 2</math>, where <math>\omega</math> can be any of the tenth [[roots of unity]]. | ||
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| + | Then the sum <math>S = a_{10} + a_{11} + \ldots = 1 + r + r^2 +\ldots = \frac{1}{1-r}</math> has value <math>\frac 1{1 - \omega / 2}</math>. Different choices of <math>\omega</math> clearly lead to different values for <math>S</math>, so we don't need to worry about the distinctness condition in the problem. Thus our final answer will be <math>X = \sum_{w^{10} = 1} \frac 1{1 - \omega/2}</math>. | ||
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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> | ||
Revision as of 09:41, 30 September 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 
