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Harvard-MIT Mathematics Tournament

Algebra Difficulty 5.0 AIME Find the answer United States

Problem:
Suppose that $\omega$ is a primitive $2007^{\text{th}}$ root of unity. Find $\left(2^{2007}-1\right) \sum_{j=1}^{2006} \frac{1}{2-\omega^{j}}$ .

For this problem only, you may express your answer in the form $m \cdot n^{k}+p$ , where $m, n, k$ , and $p$ are positive integers. Note that a number $z$ is a primitive $n^{\text{th}}$ root of unity if $z^{n}=1$ and $n$ is the smallest number amongst $k=1,2, \ldots, n$ such that $z^{k}=1$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
Answer: $2005 \cdot 2^{2006}+1$ . Note that
$\begin{aligned} & \frac{1}{z-\omega}+\cdots+\frac{1}{z-\omega^{2006}}=\frac{\sum_{j=1}^{2006} \prod_{i \neq j}\left(z-\omega^{i}\right)}{(z-\omega) \cdots\left(z-\omega^{2006}\right)} \\ & \quad=\frac{\frac{\mathrm{d}}{\mathrm{~d} z}\left[z^{2006}+z^{2005}+\cdots+1\right]}{z^{2006}+z^{2005}+\cdots+1}=\frac{2006 z^{2005}+2005 z^{2004}+\cdots+1}{z^{2006}+z^{2005}+\cdots+1} \cdot \frac{z-1}{z-1} \\ & \quad=\frac{2006 z^{2006}-z^{2005}-z^{2004}-\cdots-1}{z^{2007}-1} \cdot \frac{z-1}{z-1}=\frac{2006 z^{2007}-2007 z^{2006}+1}{\left(z^{2007}-1\right)(z-1)} . \end{aligned}$
Plugging in $z=2$ gives $\frac{2005 \cdot 2^{2006}+1}{2^{2007}-1}$ ; whence the answer.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.