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
A regular polygon with \(2n + 1\) sides has \(2n + 1\) vertices. We want to select 3 vertices to form a triangle such that the center of the polygon is inside that triangle.

### Step 2: Selection of vertices
To ensure that the center of the polygon lies inside the triangle formed by the selected vertices, we must select the vertices in such a way that they do not all lie on one side of the center. This means we cannot select three consecutive vertices.

### Step 3: Count the valid selections
1. **Total selections of vertices**: The total number of ways to choose 3 vertices from \(2n + 1\) vertices is given by the combination formula:
\[ \binom{2n + 1}{3} = \frac{(2n + 1)(2n)(2n - 1)}{6} \]

2. **Invalid selections**: We need to subtract the cases where the center does not lie inside the triangle. This occurs when all selected vertices are on one side of the center. For any selection of 3 consecutive vertices, the center will not be inside the triangle formed by them.

- If we consider a selection of 3 consecutive vertices, we can visualize this as choosing a starting vertex and then selecting the next two consecutive vertices. For \(2n + 1\) vertices, there are \(2n + 1\) ways to choose a starting vertex, leading to \(2n + 1\) invalid selections.

### Step 4: Calculate the valid triangles
Now, we can calculate the number of valid triangles:
\[ \text{Valid triangles} = \binom{2n + 1}{3} - (2n + 1) \]
Substituting the values:
\[ = \frac{(2n + 1)(2n)(2n - 1)}{6} - (2n + 1) \]

### Step 5: Simplify the expression
Factor out \((2n + 1)\):
\[ = (2n + 1) \left( \frac{(2n)(2n - 1)}{6} - 1 \right) \]
Now simplifying the term inside the parentheses:
\[ = (2n + 1) \left( \frac{2n(2n - 1) - 6}{6} \right) \]
\[ = (2n + 1) \left( \frac{4n^2 - 2n - 6}{6} \right) \]

### Step 6: Final expression
The final number of triangles that can be formed is:
\[ \frac{(2n + 1)(4n^2 - 2n - 6)}{6} \]

### Conclusion
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:
\[ \frac{(2n + 1)(n^2 + n - 1)}{6} \]