If `alpha` is an `n^(th)` roots of unity, then `1+2alpha+3alpha^(2)+……..+nalpha^(n-1)` equals

Options

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

Explanation

To solve the problem, we need to find the value of the expression \( S = 1 + 2\alpha + 3\alpha^2 + \ldots + n\alpha^{n-1} \), where \( \alpha \) is an \( n^{th} \) root of unity.

### Step-by-step Solution:

1. **Define the Sum**:
Let \( S = 1 + 2\alpha + 3\alpha^2 + \ldots + n\alpha^{n-1} \).

2. **Multiply by \( \alpha \)**:
Multiply both sides of the equation by \( \alpha \):
\[ \alpha S = \alpha + 2\alpha^2 + 3\alpha^3 + \ldots + n\alpha^n \]
Note that since \( \alpha^n = 1 \), we can replace \( n\alpha^n \) with \( n \).

3. **Rearranging the Terms**:
Now we can write:
\[ \alpha S = \alpha + 2\alpha^2 + 3\alpha^3 + \ldots + n \]
Rearranging gives:
\[ \alpha S = n + \sum_{k=1}^{n-1} k\alpha^{k+1} \]

4. **Subtracting the Two Equations**:
Now, subtract \( \alpha S \) from \( S \):
\[ S - \alpha S = 1 + (2\alpha - \alpha) + (3\alpha^2 - 2\alpha^2) + \ldots + (n\alpha^{n-1} - n\alpha^n) \]
This simplifies to:
\[ S(1 - \alpha) = 1 + \alpha + 2\alpha^2 + \ldots + (n-1)\alpha^{n-1} \]

5. **Using the Formula for the Sum of a Geometric Series**:
The right-hand side can be recognized as a geometric series. The sum of a geometric series is given by:
\[ \sum_{k=0}^{n-1} x^k = \frac{1 - x^n}{1 - x} \]
Here, we can use this to find the sum of the series.

6. **Final Expression**:
Thus, we have:
\[ S(1 - \alpha) = \frac{1 - \alpha^n}{(1 - \alpha)^2} \]
Since \( \alpha^n = 1 \) (as \( \alpha \) is an \( n^{th} \) root of unity), we can simplify this to:
\[ S(1 - \alpha) = \frac{1 - 1}{(1 - \alpha)^2} = 0 \]

7. **Solving for \( S \)**:
Therefore, we find:
\[ S = \frac{n}{1 - \alpha} \]

### Conclusion:
The final result is:
\[ S = \frac{n}{1 - \alpha} \]