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Algebra Difficulty 5.0 AIME Find the answer United States

Problem:
Suppose that ω\omega is a primitive 2007th2007^{\text{th}} root of unity. Find (22007−1)∑j=1200612−ωj\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⋅nk+pm \cdot n^{k}+p, where m,n,km, n, k, and pp are positive integers. Note that a number zz is a primitive nthn^{\text{th}} root of unity if zn=1z^{n}=1 and nn is the smallest number amongst k=1,2,…,nk=1,2, \ldots, n such that zk=1z^{k}=1.

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

Solution

Solution:
Answer: 2005⋅22006+12005 \cdot 2^{2006}+1. Note that
1z−ω+⋯+1z−ω2006=∑j=12006∏i≠j(z−ωi)(z−ω)⋯(z−ω2006)=d dz[z2006+z2005+⋯+1]z2006+z2005+⋯+1=2006z2005+2005z2004+⋯+1z2006+z2005+⋯+1⋅z−1z−1=2006z2006−z2005−z2004−⋯−1z2007−1⋅z−1z−1=2006z2007−2007z2006+1(z2007−1)(z−1). \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=2z=2 gives 2005⋅22006+122007−1\frac{2005 \cdot 2^{2006}+1}{2^{2007}-1}; whence the answer.

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