If 1, a, a^2, ldots, a^{n-1} are the nth root | vedclass

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(B) Let $\omega = a$. The $n$th roots of unity are $1, \omega, \omega^2, \ldots, \omega^{n-1}$.We know that $x^n - 1 = (x-1)(x-\omega)(x-\omega^2) \ld

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$\frac{(n-2) 2^{n-1}+1}{2^n-1}$

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(B) Let $\omega = a$. The $n$th roots of unity are $1, \omega, \omega^2, \ldots, \omega^{n-1}$.We know that $x^n - 1 = (x-1)(x-\omega)(x-\omega^2) \ldots (x-\omega^{n-1})$.Taking the natural logarithm on both sides:$\ln(x^n - 1) = \ln(x-1) + \ln(x-\omega) + \ln(x-\omega^2) + \ldots + \ln(x-\omega^{n-1})$.Differentiating with respect to $x$:$\frac{n x^{n-1}}{x^n - 1} = \frac{1}{x-1} + \sum_{i=1}^{n-1} \frac{1}{x-\omega^i}$.Rearranging to isolate the summation:$\sum_{i=1}^{n-1} \frac{1}{x-\omega^i} = \frac{n x^{n-1}}{x^n - 1} - \frac{1}{x-1}$.Setting $x = 2$:$\sum_{i=1}^{n-1} \frac{1}{2-\omega^i} = \frac{n \cdot 2^{n-1}}{2^n - 1} - \frac{1}{2-1} = \frac{n \cdot 2^{n-1}}{2^n - 1} - 1$.Simplifying the expression:$\frac{n \cdot 2^{n-1} - (2^n - 1)}{2^n - 1} = \frac{n \cdot 2^{n-1} - 2 \cdot 2^{n-1} + 1}{2^n - 1} = \frac{(n-2) 2^{n-1} + 1}{2^n - 1}$.

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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