Question

If $1,a_1,a_2...,a_{n-1}$ are $n^{th}$ roots of unity, then $\frac{1}{1-a_1}+\frac{1}{1-a_2}+...+\frac{1}{1-a_{n-1}}$ equals

• A. $\frac{2^n-1}{n}$
• B. $\frac{n-1}{2}$
• C. $\frac{n}{n-1}$
• D. None of these

Answer 1

The correct option is B $\frac{n-1}{2}$

Given $z^n=1,z=1,a_1,a_2,...,a_{n-1}$
Let $a=\frac{1}{1-z}\Rightarrow z=1-\frac{1}{a}$
$\therefore {(1-\frac{1}{a})}^n=1$
$\Rightarrow {(a-1)}^n-a^n=0$
$\Rightarrow -C_1{a}^{n-1}+C_2{a}^{n-2}+...+{(-1)}^n=0$ where $a=\frac{1}{1-a_1},\frac{1}{1-a_2}.....\frac{1}{1-a_{n-1}}$
$\Rightarrow \frac{1}{1-a_1}+\frac{1}{1-a_2}+.....+\frac{1}{1-a_{n-1}}=\frac{{}^n{c}_2}{n}=\frac{n-1}{2}$