`sin"" (pi )/( n ) + sin "" ( 3pi)/( n ) + sin"" ( 5pi)/( n ) +"…."n` terms =

Options

A. 1
B. 0
C. `n//2`
D. none

Answer

To find the sum of the series \( \sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{3\pi}{n}\right) + \sin\left(\frac{5\pi}{n}\right) + \ldots \) (up to \( n \) terms), we can use the formula for the sum of a sine series.

### Step-by-Step Solution:

1. **Identify the Series**: The series we need to sum is:
\[ S = \sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{3\pi}{n}\right) + \sin\left(\frac{5\pi}{n}\right) + \ldots + \sin\left(\frac{(2n-1)\pi}{n}\right) \]
This series has \( n \) terms.

2. **Use the Sine Sum Formula**: The formula for the sum of sine series is:
\[ S = \frac{\sin\left(\frac{n \beta}{2}\right) \cdot \sin\left(\alpha + \frac{(n-1) \beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)} \]
where \( \alpha \) is the first term and \( \beta \) is the common difference.

3. **Assign Values**: In our case:
- The first term \( \alpha = \frac{\pi}{n} \)
- The common difference \( \beta = \frac{2\pi}{n} \)

4. **Substitute into the Formula**:
\[ S = \frac{\sin\left(\frac{n \cdot \frac{2\pi}{n}}{2}\right) \cdot \sin\left(\frac{\pi}{n} + \frac{(n-1) \cdot \frac{2\pi}{n}}{2}\right)}{\sin\left(\frac{\frac{2\pi}{n}}{2}\right)} \]

5. **Simplify the Terms**:
- The numerator becomes:
\[ \sin(\pi) \cdot \sin\left(\frac{\pi}{n} + \frac{(n-1)\pi}{n}\right) = \sin(\pi) \cdot \sin\left(\frac{n\pi}{n}\right) = \sin(\pi) \cdot \sin(\pi) = 0 \]
- The denominator is:
\[ \sin\left(\frac{\pi}{n}\right) \]

6. **Final Result**:
Since the numerator is 0, the entire expression simplifies to:
\[ S = \frac{0}{\sin\left(\frac{\pi}{n}\right)} = 0 \]

Thus, the sum of the series is:
\[ \boxed{0} \]