If 1,α,α2,........,αn-1 are the nth roots of unity, then displaystyle∑i=1n-1(1/2-ai) is equal to

Question

If 1,α,α2,........,αn-1 are the nth roots of unity, then displaystyle∑i=1n-1(1/2-ai) is equal to (A) (n-2)⋅ 2n (B) ((n-2)2n-1+1/2n-1) (C) ((n-2)2n-1/2n-1) (D) none of these

Tardigrade Admin · Accepted Answer

Since 1, α, α2, ...αn-1 are the nth roots of unity ∴ xn-1=(x-1)(x-α)(x-α)2 ldots ldots(x-αn-1) ∴ log(xn-1)=log(x-1)+log(x-α) +log(x-α)2+.....+log(x-αn-1) Diff. both sides w.r.t. âxâ, we get (nxn-1/xn-1)=(1/x-1)+(1/x+α) +(1/x-α2)+...+(1/x-αn-1) Putting x = 2, we get (n2n-1/2n-1)=(1/1)+(1/2-α)+(1/2-α2)+...+(1/2-αn-1) ∴ (n.2n-1/2n-1)-1 = displaystyle ∑i=1n=1(1/2-α). Hence displaystyle ∑i=1n=1(1/2-αi) =(n.2n-1-2n+1/2n-1) =((n-2)2n-1+1/2n+1)

Q. If $1, α, α^{2}, ........, α^{n - 1}$ are the nth roots of unity, then $i = 1 \sum n - 1 \frac{1}{2 - a ^{i}}$ is equal to

$(n - 2) \cdot 2^{n}$

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

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

none of these

Q. If 1,α,α2,........,αn−1 are the nth roots of unity, then i=1∑n−1​2−ai1​ is equal to

(n−2)⋅2n

2n−1(n−2)2n−1+1​

2n−1(n−2)2n−1​

none of these

Q. If $1, α, α^{2}, ........, α^{n - 1}$ are the nth roots of unity, then $i = 1 \sum n - 1 \frac{1}{2 - a ^{i}}$ is equal to

$(n - 2) \cdot 2^{n}$

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

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