Example 6 As shown in Figure 5, a rhombus $A B C D$ with side length $n$, where the vertex angle $A$ is $60^{\circ}$. Three sets of equally spaced parallel lines, each parallel to $A B$, $A D$, and $B D$, respectively, divide the rhombus into $2 n^{2}$ equilateral triangles with side length 1. Try to find the number of trapezoids $s(n)$ formed by the line segments in the figure.
Given that any two line segments in the figure are either parallel or intersect at a $60^{\circ}$ acute angle, all trapezoids formed by the line segments in the figure are isosceles trapezoids with base angles of $60^{\circ}$.

For such trapezoids, if the intersection point of the extensions of the two non-parallel sides is inside or on the boundary of the rhombus, it is called an "internal trapezoid"; if the intersection point is outside the rhombus, it is called an "external trapezoid".
(1) Find the number of internal trapezoids $f(n)$.
A triangle with side length $k$ is called a "level-$k$ triangle," and a trapezoid with the longer base of length $k$ is called a "level-$k$ trapezoid." A level-$k$ trapezoid with leg length $r (r < k)$ is called a "type-$(k, r)$ trapezoid." Furthermore, all triangles in the figure either point upwards or downwards, referred to as "upright triangles" and "inverted triangles," respectively.

It is easy to see that each type-$(k, r)$ trapezoid can be viewed as a level-$k$ triangle with a level-$(k-r)$ triangle cut off. Each level-$k$ triangle can produce $3(k-1)$ types of level-$k$ trapezoids (type-$(k, 1)$, type-$(k, 2)$, ..., type-$(k, k-1)$ trapezoids, each three times).

Calculate the number of level-$k$ triangles.
Taking $A$ as the origin and $AB$ and $AD$ as the $x$-axis and $y$-axis, respectively, in a skew coordinate system, the horizontal coordinate of the left endpoint $P$ of the base of each level-$k$ upright triangle can take $0, 1, \cdots, n-k$, a total of $n-k+1$ values, and the vertical coordinate of $P$ can also take $0, 1, \cdots, n-k$, a total of $n-k+1$ values.
Therefore, there are $(n-k+1)^2$ level-$k$ upright triangles.
By symmetry, there are also $(n-k+1)^2$ level-$k$ inverted triangles.
Thus, there are $2(n-k+1)^2$ level-$k$ triangles.
Hence, there are $6[n-(k-1)]^2(k-1)$ level-$k$ internal trapezoids. Summing up, we get
$$
\begin{array}{l}
f(n)=6 \sum_{k=1}^{n}[n-(k-1)]^{2}(k-1) \\
=6 \sum_{i=1}^{n-1}(n-i)^{2} i=6 \sum_{j=1}^{n-1} j^{2}(n-j) \\
=6 n \sum_{j=1}^{n-1} j^{2}-6 \sum_{j=1}^{n-1} j^{3} \\
=6 n \cdot \frac{n(n-1)(2 n-1)}{6}-6\left[\frac{n(n-1)}{2}\right]^{2} \\
=\frac{n^{2}\left(n^{2}-1\right)}{2} .
\end{array}
$$
(2) Find the number of external trapezoids $g(n)$.

First, consider the case where the external intersection point is outside the segment $AB$. As shown in Figure 6, take any $i, j$ such that $1 \leqslant j < i \leqslant n$.
Figure 6

Let the skew coordinates of the points be
$T_{i}(i, 0), P_{j}(0, j), Q_{j}(j, 0), R_{i j}(i, j)$.
Extend $P_{j} Q_{j}$ to intersect the line $x=i$ at point $M_{i}$, and there are $n-j$ such intersection points on the extension of $P_{j} Q_{j}$.

For a fixed $j$, $\triangle M_{i j} P_{j} R_{i j}$ is an inverted triangle. As point $T_{i}$ moves along $AB$, point $R_{i i}$ moves along the line $y=j$.

Since $P_{j} R_{i j} \parallel AB$, there are $j+1$ parallel lines between these two segments (including the two segments themselves). Any two of these parallel lines intersect $\triangle M_{i j} P_{j} R_{i j}$ to form a trapezoid.

Thus, there are $\mathrm{C}_{j+1}^{2}=\frac{j(j+1)}{2}$ such trapezoids, all with $M_{i j}$ as the external intersection point. There are $n-j$ such external intersection points on the extension of $P_{j} Q_{j}$, so when $j$ is fixed, there are $\frac{j(j+1)}{2}(n-j)$ external trapezoids.

Let $j$ take all values from $1, 2, \cdots, n-1$, then all external intersection points outside $AB$ form $\sum_{j=1}^{n-1} \frac{j(j+1)}{2}(n-j)$ external trapezoids.

By symmetry, the external intersection points outside the other three sides of the rhombus also form the same number of external trapezoids. Thus, the total number of external trapezoids is
$$
\begin{array}{l}
g(n)=4 \sum_{j=1}^{n-1} \frac{j(j+1)}{2}(n-j) \\
=2 \sum_{j=1}^{n-1}\left[(n-1) j^{2}+n j-j^{3}\right] \\
=2(n-1) \cdot \frac{n(n-1)(2 n-1)}{6}+ \\
2 n \cdot \frac{n(n-1)}{2}-2\left[\frac{n(n-1)}{2}\right]^{2} \\
=\frac{n(n+2)\left(n^{2}-1\right)}{6} . \\
\text { Hence } s(n)=f(n)+g(n) \\
=\frac{n^{2}\left(n^{2}-1\right)}{2}+\frac{n(n+2)\left(n^{2}-1\right)}{6} \\
=\frac{n\left(n^{2}-1\right)(2 n+1)}{3} .
\end{array}
$$