User talk:JBL

For later use (Mock AIME 2006)

Let the ratio of consecutive terms of the sequence be $r \in \mathbb{C}$ . Then we have by the given that $1 = a_{10} = r^{10} a_0 = 1024r^{10}$ so $r^{10} = 2^{-10}$ and $r = \frac \omega 2$ , where $\omega$ can be any of the tenth roots of unity.

Then the sum $S = a_{10} + a_{11} + \ldots = 1 + r + r^2 +\ldots = \frac{1}{1-r}$ has value $\frac 1{1 - \omega / 2}$ . Different choices of $\omega$ clearly lead to different values for $S$ , so we don't need to worry about the distinctness condition in the problem. Thus our final answer will be $X = \sum_{w^{10} = 1} \frac 1{1 - \omega/2}$ .

For every choice of $\omega$ , $\overline \omega$ is also a 10th root of unity, so if we take 2 copies of $X$ we can pair up terms to get $2X = \sum_{\omega^{10} = 1} \frac 1{1 - \omega/2} + \frac1{1 - \overline{\omega}/2} = \sum_{\omega^{10} = 1} \frac {(1 - \omega/2) + (1 - \overline{\omega}/2)}{(1 - \omega/2)(1 - \overline{\omega}/2)}$

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User talk:JBL

For later use (Mock AIME 2006)