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If $1,a,a^2,...,a^{n-1}$ are the n$^\text{th}$ roots of unity, then prove that$$\sum_{i=1}^{n-1}\frac{1}{2-a^i}=\frac{(n-2)2^{n-1}+1}{2^n-1}$$

$$ \alpha_r=e^{i\tfrac{2\pi r}{n}}=a^{r-1}\\ x^n=1\implies x^n-1=(x-1)(x-a)(x-a^2)...(x-a^{n-1})=0 $$

$$ \sum_{i=1}^{n-1}\frac{1}{2-a^i}=\frac{1}{2-a}+\frac{1}{2-a^2}+\frac{1}{2-a^3}+....+\frac{1}{2-a^{n-1}}\\ = $$ Note: This is solved using derivative of logarithm in Problem based on sum of reciprocal of 𝑛𝑡ℎ roots of unity which is good, but I am looking for more of a direct and easier way to find the solution.

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marked as duplicate by Martin R, José Carlos Santos, Kemono Chen, YiFan, Shuhao Cao Feb 22 at 23:55

This question was marked as an exact duplicate of an existing question.

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Let $x_i=\frac{1}{2-a_i}$.

Thus, $$a_i=2-\frac{1}{x_i},$$ which says that $x_i$ they are roots of the equation: $$\left(2-\frac{1}{x_i}\right)^n-1=0$$ or $$(2x_i-1)^n-x_i^n=0$$ or $$(2^n-1)x_i^n-n2^{n-1}x_i^{n-1}+...=0,$$ which gives $$x_1+x_2+...+x_n=\frac{n2^{n-1}}{2^n-1}.$$ Thus, $$\sum_{i=1}^{n-1}\frac{1}{2-a^i}=\frac{n2^{n-1}}{2^n-1}-1=\frac{(n-2)2^{n-1}+1}{2^n-1}.$$

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I'll do $\sum_{i=0}^{n-1}1/(2-a^i)$ instead: \begin{align}\sum_{i=0}^{n-1}\frac1{2-a^i}&=\frac12\sum_{i=0}^{n-1}\frac1{1-a^i/2} =\frac12\sum_{i=0}^{n-1}\sum_{k=0}^\infty\frac{a^{ik}}{2^k} =\frac12\sum_{k=0}^\infty\sum_{i=0}^{n-1}\frac{a^{ik}}{2^k}\\ &=\frac 12\sum_{l=0}^\infty\frac{n}{2^{nl}}=\frac{n}{2(1-2^{-n})} =\frac{n2^{n-1}}{2^n-1}. \end{align}

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  • $\begingroup$ i am having trouble understand ur method. could u pls explain each steps ? $\endgroup$ – ss1729 Mar 3 at 12:47

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