If 1, ω, ldots, ωn-1 are the nth roots of unity, then value of (1/2-ω)+(1/2-ω2)+⋅s+(1/2-ωn-1) equals

Question

If 1, ω, ldots, ωn-1 are the nth roots of unity, then value of (1/2-ω)+(1/2-ω2)+⋅s+(1/2-ωn-1) equals (A) (1/2n-1) (B) (n(2n-1)/2n+1) (C) ((n-2) 2n-1/2n-1) (D) none of these

Tardigrade Admin · Accepted Answer

We know that (1/x-1)+(1/x-ω)+(1/x-ω2)+⋅s+(1/x-ωn-1)=(n(xn-1)/xn-1) Putting x=2, we get (1/2-ω)+(1/2-ω2)+⋅s+(1/2-ωn-1)=(n(2n-1)/2n-1)

Q. If $1, ω, \dots, ω^{n - 1}$ are the $n$ th roots of unity, then value of $\frac{1}{2 - ω} + \frac{1}{2 - ω ^{2}} + \dots + \frac{1}{2 - ω ^{n - 1}}$ equals

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

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

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

none of these

We know that
$\frac{1}{x - 1} + \frac{1}{x - ω} + \frac{1}{x - ω ^{2}} + \dots + \frac{1}{x - ω ^{n - 1}} = \frac{n ( x ^{n - 1} )}{x ^{n} - 1}$
Putting $x = 2$ , we get
$\frac{1}{2 - ω} + \frac{1}{2 - ω ^{2}} + \dots + \frac{1}{2 - ω ^{n - 1}} = \frac{n ( 2 ^{n - 1} )}{2 ^{n} - 1}$

Q. If 1,ω,…,ωn−1 are the nth roots of unity, then value of 2−ω1​+2−ω21​+⋯+2−ωn−11​ equals

2n−11​

2n+1n(2n−1)​

2n−1(n−2)2n−1​