The number of triangles that can be formed form a regular polygon of 2n + 1 sides such that the centre of the polygon lies inside the triangle is

Options

A. `(n)/(2) (n+1)`
B. `(n^(2)+3n)/(2)`
C. `(n(n+1)(2n+1))/(6)`
D. `((n(n+1))/(2))^(2)`

Answer

To solve the problem of finding the number of triangles that can be formed from a regular polygon of \(2n + 1\) sides such that the center of the polygon lies inside the triangle, we can follow these steps:

### Step 1: Understand the Problem
We have a regular polygon with \(2n + 1\) sides. We need to form triangles from the vertices of this polygon such that the center of the polygon is inside the triangle.

### Step 2: Total Number of Triangles
The total number of triangles that can be formed from \(2n + 1\) vertices is given by the combination formula:
\[ \text{Total triangles} = \binom{2n + 1}{3} \]

### Step 3: Identify Invalid Triangles
A triangle will not contain the center of the polygon if it includes three consecutive vertices. Therefore, we need to count the number of triangles that do not satisfy our condition.

1. **Count the Invalid Triangles**:
- If we select three consecutive vertices, we can choose any vertex as the starting point. Since the polygon has \(2n + 1\) vertices, there are \(2n + 1\) ways to choose three consecutive vertices.

### Step 4: Calculate Valid Triangles
Now, we can find the number of valid triangles by subtracting the number of invalid triangles from the total number of triangles:
\[ \text{Valid triangles} = \text{Total triangles} - \text{Invalid triangles} \]
\[ \text{Valid triangles} = \binom{2n + 1}{3} - (2n + 1) \]

### Step 5: Simplify the Expression
Now we will simplify the expression:
1. Calculate \(\binom{2n + 1}{3}\):
\[ \binom{2n + 1}{3} = \frac{(2n + 1)(2n)(2n - 1)}{6} \]

2. Substitute this back into the valid triangles equation:
\[ \text{Valid triangles} = \frac{(2n + 1)(2n)(2n - 1)}{6} - (2n + 1) \]

3. Factor out \(2n + 1\):
\[ \text{Valid triangles} = (2n + 1) \left( \frac{(2n)(2n - 1)}{6} - 1 \right) \]

4. Simplify the expression inside the parentheses:
\[ \frac{(2n)(2n - 1)}{6} - 1 = \frac{2n(2n - 1) - 6}{6} = \frac{4n^2 - 2n - 6}{6} \]

5. Therefore, the final expression for the number of valid triangles is:
\[ \text{Valid triangles} = \frac{(2n + 1)(4n^2 - 2n - 6)}{6} \]

### Final Answer
Thus, the number of triangles that can be formed from a regular polygon of \(2n + 1\) sides such that the center of the polygon lies inside the triangle is:
\[ \frac{(2n + 1)(4n^2 - 2n - 6)}{6} \]