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AP EAMCET · Maths · Complex Number

If \(1, a, a^2, \ldots, a^{n-1}\) are the \(n\)th roots of unity. Then \(\sum_{i=1}^{n-1} \frac{1}{2-a^i}\) is equal to

  1. A \((n-2) 2^n\)
  2. B \(\frac{(n-2) 2^{n-1}+1}{2^n-1}\)
  3. C \(\frac{(n-2) 2^{n-1}}{2^n-1}\)
  4. D \(\frac{1}{(n-2) 2^n}\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\sum_{i=1}^{n-1} \frac{1}{2-\alpha^i}\) \(1, \alpha, \alpha^2, \ldots \ldots, \alpha^{n-1}\) are the \(n\)th root of unity…