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

Answer: (B)

Given $z^n = 1, z = 1, a_1, a_2 , .... , a_{n -1} \, ...(i)$
Let $\alpha = \frac{1}{ 1 -z} $
$\therefore z = 1 - \frac{1}{\alpha}$
$\therefore \left( 1 - \frac{1}{\alpha}\right)^n = 1$ (By (i))
$\Rightarrow (\alpha - 1)^n - \alpha^n = 0$
$\Rightarrow -C_1 \,\alpha^{n - 1} + C_2 \alpha^{n - 2} + ... + (-)^n = 0$
where $\alpha = \frac{1}{1 - a_1}, \frac{1}{1 - a_2}, ... , \frac{1}{1 - a_{n - 1}}$
$\Rightarrow \frac{1}{ 1 - a_1} + \frac{1}{a - a_2} + ... + \frac{1}{ 1 - a_{n - 1}} $
$= \frac{\,{}^nC_2}{n} = \frac{n - 1}{2}$