`underset(r=0)overset(n)(sum)sin^(2)""(rpi)/(n)` is equal to

Explanation

To solve the problem, we need to evaluate the summation:

\[ \sum_{r=0}^{n} \sin^2\left(\frac{r \pi}{n}\right) \]

### Step 1: Use the identity for \(\sin^2\)
We can use the trigonometric identity:

\[ \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \]

Applying this identity to our summation, we have:

\[ \sin^2\left(\frac{r \pi}{n}\right) = \frac{1 - \cos\left(\frac{2r \pi}{n}\right)}{2} \]

### Step 2: Substitute into the summation
Substituting this into the summation gives:

\[ \sum_{r=0}^{n} \sin^2\left(\frac{r \pi}{n}\right) = \sum_{r=0}^{n} \frac{1 - \cos\left(\frac{2r \pi}{n}\right)}{2} \]

This can be split into two separate summations:

\[ = \frac{1}{2} \sum_{r=0}^{n} 1 - \frac{1}{2} \sum_{r=0}^{n} \cos\left(\frac{2r \pi}{n}\right) \]

### Step 3: Evaluate the first summation
The first summation is straightforward:

\[ \sum_{r=0}^{n} 1 = n + 1 \]

Thus, we have:

\[ \frac{1}{2} (n + 1) \]

### Step 4: Evaluate the second summation
Now, we need to evaluate:

\[ \sum_{r=0}^{n} \cos\left(\frac{2r \pi}{n}\right) \]

This is a geometric series. The sum of cosines can be evaluated as follows:

\[ \sum_{r=0}^{n} \cos\left(\frac{2r \pi}{n}\right) = \frac{\sin\left(\frac{(n + 1) \pi}{n}\right)}{\sin\left(\frac{\pi}{n}\right)} \]

Since \(\sin\left(\frac{(n + 1) \pi}{n}\right) = \sin(\pi) = 0\), we find that:

\[ \sum_{r=0}^{n} \cos\left(\frac{2r \pi}{n}\right) = 0 \]

### Step 5: Combine results
Putting this back into our expression, we have:

\[ \sum_{r=0}^{n} \sin^2\left(\frac{r \pi}{n}\right) = \frac{1}{2}(n + 1) - \frac{1}{2}(0) = \frac{n + 1}{2} \]

### Final Answer
Thus, the final result is:

\[ \sum_{r=0}^{n} \sin^2\left(\frac{r \pi}{n}\right) = \frac{n + 1}{2} \]

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