If ` beta ne 1 ` be any nth root of unity, then ` 1 + 3 beta + 5 beta ^(2) + . . . n` terms equals

Options

A. `(2 n)/( 1 - beta)`
B. ` - (2 beta)/( 1 - beta)`
C. ` - (2 n)/( ( 1- beta)^(2))`
D. none

Answer

To solve the problem, we need to find the sum of the series \( S = 1 + 3\beta + 5\beta^2 + \ldots + (2n - 1)\beta^{n-1} \), where \( \beta \) is an nth root of unity and \( \beta \neq 1 \).

### Step-by-step Solution:

1. **Define the Series**:
Let \( S = 1 + 3\beta + 5\beta^2 + \ldots + (2n - 1)\beta^{n-1} \).

2. **Multiply by \( \beta \)**:
Multiply the entire series by \( \beta \):
\[ \beta S = \beta + 3\beta^2 + 5\beta^3 + \ldots + (2n - 1)\beta^n \]

3. **Rearranging Terms**:
Notice that the last term in \( \beta S \) is \( (2n - 1)\beta^n \). Since \( \beta^n = 1 \) for any nth root of unity, we can rewrite it as:
\[ \beta S = \beta + 3\beta^2 + 5\beta^3 + \ldots + (2n - 1) \]

4. **Subtracting the Two Equations**:
Now subtract \( \beta S \) from \( S \):
\[ S - \beta S = (1 - \beta) + (3 - 3\beta)\beta + (5 - 5\beta^2)\beta^2 + \ldots + (2n - 1 - (2n - 1)\beta^{n-1}) \]
This simplifies to:
\[ S(1 - \beta) = 1 + 2\beta + 2\beta^2 + \ldots + 2\beta^{n-1} \]

5. **Recognizing a Geometric Series**:
The right-hand side is a geometric series with first term \( 1 \) and common ratio \( \beta \):
\[ 1 + 2\beta + 2\beta^2 + \ldots + 2\beta^{n-1} = 1 + 2\left(\frac{1 - \beta^{n}}{1 - \beta}\right) \]
Since \( \beta^n = 1 \):
\[ = 1 + 2\left(\frac{1 - 1}{1 - \beta}\right) = 1 \]

6. **Final Equation**:
Thus, we have:
\[ S(1 - \beta) = 1 \]
Therefore,
\[ S = \frac{1}{1 - \beta} \]

### Conclusion:
The sum of the series \( S = 1 + 3\beta + 5\beta^2 + \ldots + (2n - 1)\beta^{n-1} \) is given by:
\[ S = \frac{1}{1 - \beta} \]