If $1, a, a^2, \ldots, a^{n-1}$ are the $n$th roots of unity. Then $\sum_{i=1}^{n-1} \frac{1}{2-a^i}$ is equal to

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Aicharya · Accepted Answer

$\sum_{i=1}^{n-1} \frac{1}{2-\alpha^i}$ $1, \alpha, \alpha^2, \ldots \ldots, \alpha^{n-1}$ are the $n$th root of unity $\begin{array}{ll} \because & x^n-1=0 \ \Rightarrow & x^n-1=(x-1)(x-\alpha)\left(x-\alpha^2\right) \ldots \ldots\left(x-\alpha^{n-1}\right) \ \Rightarrow \quad & \log \left(x^n-1\right)=\log (x-1)+\log (x-\alpha) \ & \quad+\ldots . . \log \left(x-\alpha^{n-1}\right) \end{array}$ Differentiating w.r.t ' $x$ ', we get $\begin{aligned} & \Rightarrow \quad \frac{n x^{n-1}}{x^n-1}=\frac{1}{(x-1)}+\frac{1}{(x-\alpha)}+\ldots \ldots+\frac{1}{\left(x-\alpha^{n-1}\right)} \ & \text {At }(x=2) \ & \Rightarrow \quad \frac{n \cdot 2^{n-1}}{2^n-1}=1+\frac{1}{2-\alpha}+\frac{1}{2-\alpha^2}+\ldots . .+\frac{1}{2-\alpha^{n-1}} \ & \Rightarrow \quad \sum_{i=1}^{n-1} \frac{1}{2-\alpha^i}=\left(\frac{n \cdot 2^{n-1}}{2^n-1}-1\right)=\frac{(n-2) 2^{n-1}+1}{2^n-1} \end{aligned}$

If \(1, a, a^2, \ldots, a^{n-1}\) are the \(n\)th roots of unity. Then \(\sum_{i=1}^{n-1} \frac{1}{2-a^i}\)…

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