Three. (50 points) A rhombus $ABCD$ with side length $n$ has its apex angle $\angle A$ as $60^{\circ}$. Using three sets of equally spaced parallel lines, each parallel to $AB$, $AD$, and $BD$ respectively, the rhombus is divided into $2n^2$ equilateral triangles with side length 1 (as shown in Figure 3). Try to find the number of trapezoids $s(n)$ formed by the line segments in the figure.
Three, since any two line segments in Figure 3 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) First, 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 "level-$(k, r)$ trapezoid." 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 level-$(k, r)$ trapezoid can be viewed as a level-$k$ triangle with a level-$(k-r)$ triangle cut off. Each level-$k$ triangle can form $3(k-1)$ types of level-$k$ trapezoids (where the level-$(k, i) (i=1,2, \cdots, k-1)$ trapezoids each have 3).

Now, calculate the number of level-$k$ triangles: take $A$ as the origin and $AB$, $AD$ as the $x$-axis and $y$-axis, respectively, to establish a skew coordinate system. For each level-$k$ upright triangle, the left endpoint $P$ of the base can have $0,1, \cdots, n-k$ for a total of $n-k+1$ values for the $x$-coordinate, and $0,1, \cdots, n-k$ for a total of $n-k+1$ values for the $y$-coordinate. 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. Therefore, the number of level-$k$ internal trapezoids is $6[n-(k-1)]^2(k-1)$. Summing up, we get
$$
\begin{array}{l}
f(n)=6 \sum_{k=1}^{n}[n-(k-1)]^{2}(k-1) \\
=6 \sum_{i=1}^{n-1}(n-i)^{2} i=6 \sum_{j=1}^{n-1} j^{2}(n-j) \\
=6 n \sum_{j=1}^{n-1} j^{2}-6 \sum_{j=1}^{n-1} j^{3}=\frac{n^{2}\left(n^{2}-1\right)}{2} .
\end{array}
$$
(2) Next, find the number of external trapezoids $g(n)$.

First, consider the case where the external intersection point is outside the line segment $AB$. As shown in Figure 10, for any $i, j$ such that
$1 \leqslant j < i \leqslant n$, let the skew coordinates of the points be: $T_{i}(i, 0)$, $P_{j}(0, j)$, $Q_{j}(j, 0)$, $R_{ij}(i, j)$. Extend $P_{j}Q_{j}$ to intersect the line $x=i$ at point $M_{ij}$, and there are $n-j$ points on the extension of $P_{j}Q_{j}$. For a fixed $j$, $\triangle M_{ij}P_{j}R_{ij}$ is an inverted equilateral triangle. As point $T_{i}$ moves along $AB$, point $R_{i}$ moves along the line $y=j$. Since $P_{j}R_{ij} \parallel AB$, there are $j+1$ parallel lines between these two lines (including the two lines themselves). Any two of these parallel lines form a trapezoid within $\triangle M_{ij}P_{j}R_{ij}$. Therefore, there are $\mathrm{C}_{j+1}^{2}=\frac{j(j+1)}{2}$ such trapezoids, all with $M_{ij}$ as the external intersection point. Since there are $n-j$ points on the extension of $P_{j}Q_{j}$ outside $AB$, when $j$ is fixed, a total of $\frac{j(j+1)}{2}(n-j)$ external trapezoids are formed. Now, let $j$ take all values from $1,2, \cdots, n-1$, so the total number of external trapezoids with external intersection points outside $AB$ is $\sum_{j=1}^{n-1} \frac{j(j+1)}{2}(n-j)$.

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