If `1,alpha,alpha^(2),.......,alpha^(n-1)` are the `n^(th)` roots of unity, then `sum_(i=1)^(n-1)(1)/(2-alpha^(i))` is equal to:

Options

A. `((n-2)2^(n-1)+1)/(2^(n)-1)`
B. `(n-2)2^(n)`
C. `((n-2).2^(n-1))/(2^(n)-1)`
D. None of these

Answer

To solve the problem, we need to evaluate the sum:

\[ S = \sum_{i=1}^{n-1} \frac{1}{2 - \alpha^i} \]

where \( \alpha \) is an \( n \)-th root of unity. The \( n \)-th roots of unity are given by:

\[ 1, \alpha, \alpha^2, \ldots, \alpha^{n-1} \]

and they satisfy the equation \( x^n - 1 = 0 \).

### Step 1: Understanding the Roots of Unity
The \( n \)-th roots of unity can be expressed as:

\[ \alpha = e^{2\pi i k/n} \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]

### Step 2: Using the Logarithmic Derivative
We can use the property of logarithmic differentiation on the polynomial whose roots are the \( n \)-th roots of unity:

\[ P(x) = x^n - 1 = (x - 1)(x - \alpha)(x - \alpha^2) \cdots (x - \alpha^{n-1}) \]

Taking the logarithm, we have:

\[ \log P(x) = n \log x - \sum_{i=0}^{n-1} \log(x - \alpha^i) \]

### Step 3: Differentiate with Respect to \( x \)
Differentiating both sides with respect to \( x \):

\[ \frac{P'(x)}{P(x)} = \frac{n}{x} - \sum_{i=0}^{n-1} \frac{1}{x - \alpha^i} \]

### Step 4: Evaluating at \( x = 2 \)
Substituting \( x = 2 \):

\[ \frac{P'(2)}{P(2)} = \frac{n}{2} - \sum_{i=0}^{n-1} \frac{1}{2 - \alpha^i} \]

### Step 5: Calculate \( P(2) \) and \( P'(2) \)
Calculating \( P(2) \):

\[ P(2) = 2^n - 1 \]

Calculating \( P'(x) \):

\[ P'(x) = n x^{n-1} \]

Thus,

\[ P'(2) = n \cdot 2^{n-1} \]

### Step 6: Substitute Back
Now substituting back into the equation:

\[ \frac{n \cdot 2^{n-1}}{2^n - 1} = \frac{n}{2} - \sum_{i=0}^{n-1} \frac{1}{2 - \alpha^i} \]

Rearranging gives:

\[ \sum_{i=0}^{n-1} \frac{1}{2 - \alpha^i} = \frac{n}{2} - \frac{n \cdot 2^{n-1}}{2^n - 1} \]

### Step 7: Simplifying the Expression
Now we can simplify the expression for \( S \):

\[ S = \sum_{i=1}^{n-1} \frac{1}{2 - \alpha^i} = \frac{n \cdot 2^{n-1}}{2^n - 1} - \frac{1}{2 - 1} \]

### Final Result
Thus, the final result for the sum is:

\[ S = \frac{n \cdot 2^{n-1} - (2^n - 1)}{2^n - 1} \]