Given a regular $2 n+1$-sided polygon. How many triangles are there whose vertices are among the vertices of the polygon and the triangle contains the center of the polygon?
In the solution, "triangle" refers to those triangles whose vertices are among the vertices of the given polygon.

The number of all triangles is exactly the number of different ways to choose three out of the $2 n+1$ vertices, which is $\binom{2 n+1}{3}=\frac{(2 n+1) 2 n(2 n-1)}{6}$. From this, we subtract the number of triangles that do not contain the center. Let's consider such a triangle and label its vertices as $A, B, C$ in a positive rotational direction, with the longest side being $A C$. This can clearly be done uniquely. Let's examine how many ways we can choose such an $A B C$ triangle! The vertex $A$ can be placed in $(2 n+1)$ positions. The vertices $B$ and $C$ can only fall on the right side of the line connecting vertex $A$ to the center of the polygon, and these can be chosen from the $n$ vertices on that side in $\binom{n}{2}$ ways. Therefore, the number of triangles that do not contain the center is $(2 n+1)\binom{n}{2}$. According to the problem's condition, the number of triangles that meet the criteria is $\binom{2 n+1}{3}-(2 n+1)\binom{n}{2}=\frac{n(n+1)(2 n+1)}{6}$.