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 trigonometric identities and properties. Let's break it down step by step.

### Step 1: Rewrite the terms using trigonometric identities
We know that:
\[ \sqrt{1 + \cos x} = \sqrt{2} \cos\left(\frac{x}{2}\right) \]
Using this identity, we can rewrite each term in the series:
\[ S = \sqrt{1 + \cos \alpha} + \sqrt{1 + \cos 2\alpha} + \sqrt{1 + \cos 3\alpha} + \ldots + \sqrt{1 + \cos n\alpha \]
\[ = \sqrt{2} \left( \cos\left(\frac{\alpha}{2}\right) + \cos\left(\frac{2\alpha}{2}\right) + \cos\left(\frac{3\alpha}{2}\right) + \ldots + \cos\left(\frac{n\alpha}{2}\right) \right) \]

### Step 2: Sum the cosine terms
The sum of cosines can be expressed using the formula:
\[ \sum_{k=0}^{n-1} \cos(a + kd) = \frac{\sin\left(\frac{nd}{2}\right) \cos\left(a + \frac{(n-1)d}{2}\right)}{\sin\left(\frac{d}{2}\right)} \]
In our case, \( a = \frac{\alpha}{2} \) and \( d = \frac{\alpha}{2} \). Thus, we can write:
\[ \sum_{k=1}^{n} \cos\left(\frac{k\alpha}{2}\right) = \frac{\sin\left(\frac{n\alpha}{4}\right) \cos\left(\frac{\alpha}{4} + \frac{(n-1)\alpha}{4}\right)}{\sin\left(\frac{\alpha}{4}\right)} \]

### Step 3: Substitute back into the series
Now substituting back into our expression for \( S \):
\[ S = \sqrt{2} \cdot \frac{\sin\left(\frac{n\alpha}{4}\right) \cos\left(\frac{(n+1)\alpha}{4}\right)}{\sin\left(\frac{\alpha}{4}\right)} \]

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