If 1, a1, a2, ...., an - 1 are nth roots of unity, then (1/ 1 - a1) +(1/1 - a2) + ... + (1/ 1 - an - 1) equals

Question

If 1, a1, a2, ...., an - 1 are nth roots of unity, then (1/ 1 - a1) +(1/1 - a2) + ... + (1/ 1 - an - 1) equals (A) (2n - 1/n)  (B) (n - 1/2) (C) (n/n- 1) (D) None of these

Tardigrade Admin · Accepted Answer

Given zn = 1, z = 1, a1, a2 , .... , an -1 ...(i) Let α = (1/ 1 -z) ∴ z = 1 - (1/α) ∴ ( 1 - (1/α))n = 1 (By (i)) ⇒ (α - 1)n - αn = 0 ⇒ -C1 αn - 1 + C2 αn - 2 + ... + (-)n = 0 where α = (1/1 - a1), (1/1 - a2), ... , (1/1 - an - 1) ⇒ (1/ 1 - a1) + (1/a - a2) + ... + (1/ 1 - an - 1) <bR> = ( nC2/n) = (n - 1/2)

Q. If $1, a_{1}, a_{2}, ...., a_{n - 1}$ are $n^{t h}$ roots of unity, then $\frac{1}{1 - a _{1}} + \frac{1}{1 - a _{2}} + ... + \frac{1}{1 - a _{n - 1}}$ equals

$\frac{2 ^{n} - 1}{n}$

$\frac{n - 1}{2}$

$\frac{n}{n - 1}$

None of these

Q. If 1,a1​,a2​,....,an−1​ are nth roots of unity, then 1−a1​1​+1−a2​1​+...+1−an−1​1​ equals

n2n−1​

2n−1​

n−1n​

None of these

Q. If $1, a_{1}, a_{2}, ...., a_{n - 1}$ are $n^{t h}$ roots of unity, then $\frac{1}{1 - a _{1}} + \frac{1}{1 - a _{2}} + ... + \frac{1}{1 - a _{n - 1}}$ equals

$\frac{2 ^{n} - 1}{n}$

$\frac{n - 1}{2}$

$\frac{n}{n - 1}$