Question

If $1, \omega, \ldots, \omega^{n-1}$ are the $n$th roots of unity, then value of $\frac{1}{2-\omega}+\frac{1}{2-\omega^2}+\cdots+\frac{1}{2-\omega^{n-1}}$ equals

• A. $\frac{1}{2^n-1}$
• B. $\frac{n\left(2^n-1\right)}{2^n+1}$
• C. $\frac{(n-2) 2^{n-1}}{2^n-1}$
• D. none of these

Answer 1

Answer: (D)

We know that
$\frac{1}{x-1}+\frac{1}{x-\omega}+\frac{1}{x-\omega^2}+\cdots+\frac{1}{x-\omega^{n-1}}=\frac{n\left(x^{n-1}\right)}{x^n-1}$
Putting $x=2$, we get
$\frac{1}{2-\omega}+\frac{1}{2-\omega^2}+\cdots+\frac{1}{2-\omega^{n-1}}=\frac{n\left(2^{n-1}\right)}{2^n-1}$