Sum the series: `sqrt(1+cos alpha)+sqrt(1+cos 2alpha)+sqrt(1+cos 3alpha)` +.......to n terms, where `0ltalphaltpi`

Explanation

To sum the series \( S = \sqrt{1 + \cos \alpha} + \sqrt{1 + \cos 2\alpha} + \sqrt{1 + \cos 3\alpha} + \ldots + \sqrt{1 + \cos n\alpha} \), we can use the trigonometric identity for \( 1 + \cos x \).

### Step-by-Step Solution:

1. **Use the Trigonometric Identity**:
We know that:
\[ 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \]
Therefore,
\[ \sqrt{1 + \cos k\alpha} = \sqrt{2 \cos^2\left(\frac{k\alpha}{2}\right)} = \sqrt{2} \cos\left(\frac{k\alpha}{2}\right) \]

2. **Rewrite the Series**:
Substitute the identity into the series:
\[ S = \sqrt{2} \left( \cos\left(\frac{\alpha}{2}\right) + \cos\left(\frac{2\alpha}{2}\right) + \cos\left(\frac{3\alpha}{2}\right) + \ldots + \cos\left(\frac{n\alpha}{2}\right) \right) \]
This simplifies to:
\[ S = \sqrt{2} \left( \cos\left(\frac{\alpha}{2}\right) + \cos\left(\alpha\right) + \cos\left(\frac{3\alpha}{2}\right) + \ldots + \cos\left(\frac{n\alpha}{2}\right) \right) \]

3. **Sum the Cosine Terms**:
The sum of cosines can be expressed using the formula for the sum of cosines:
\[ \sum_{k=0}^{n-1} \cos(a + kd) = \frac{\sin\left(\frac{nd}{2}\right) \cos\left(a + \frac{(n-1)d}{2}\right)}{\sin\left(\frac{d}{2}\right)} \]
Here, let \( a = \frac{\alpha}{2} \) and \( d = \frac{\alpha}{2} \):
\[ S = \sqrt{2} \cdot \frac{\sin\left(\frac{n\alpha}{4}\right) \cos\left(\frac{\alpha}{2} + \frac{(n-1)\alpha}{4}\right)}{\sin\left(\frac{\alpha}{4}\right)} \]

4. **Final Expression**:
Thus, the final expression for the sum \( S \) becomes:
\[ S = \sqrt{2} \cdot \frac{\sin\left(\frac{n\alpha}{4}\right) \cos\left(\frac{(n+1)\alpha}{4}\right)}{\sin\left(\frac{\alpha}{4}\right)} \]

### Final Answer:
The sum of the series is:
\[ S = \sqrt{2} \cdot \frac{\sin\left(\frac{n\alpha}{4}\right) \cos\left(\frac{(n+1)\alpha}{4}\right)}{\sin\left(\frac{\alpha}{4}\right)} \]