If ` alpha ` is an n th root of unity other than unity itself, then the value of ` 1 + alpha + alpha^(2) + . . . + alpha ^(n - 1)` is

Options

A. 0
B. 1
C. `-1`
D. none of these

Answer

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

### Step-by-Step Solution:

1. **Understanding Roots of Unity**:
- The \( n \)-th roots of unity are the solutions to the equation \( z^n = 1 \). These roots can be expressed as \( 1, \alpha, \alpha^2, \ldots, \alpha^{n-1} \), where \( \alpha = e^{2\pi i k/n} \) for \( k = 0, 1, 2, \ldots, n-1 \).
- Since we are considering \( \alpha \) as an \( n \)-th root of unity other than unity itself, we have \( \alpha \neq 1 \).

2. **Using the Formula for the Sum of a Geometric Series**:
- The sum \( S \) can be recognized as a geometric series. The formula for the sum of a geometric series is:
\[ S = \frac{a(1 - r^n)}{1 - r} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.
- In our case, \( a = 1 \), \( r = \alpha \), and there are \( n \) terms. Thus, we can write:
\[ S = \frac{1(1 - \alpha^n)}{1 - \alpha} \]

3. **Evaluating \( \alpha^n \)**:
- Since \( \alpha \) is an \( n \)-th root of unity, we have \( \alpha^n = 1 \).
- Therefore, substituting this into our equation for \( S \):
\[ S = \frac{1 - 1}{1 - \alpha} = \frac{0}{1 - \alpha} = 0 \]

4. **Conclusion**:
- The value of the sum \( 1 + \alpha + \alpha^2 + \ldots + \alpha^{n-1} \) is \( 0 \).

### Final Answer:
\[ S = 0 \]