II. (Full marks 50 points) There are $n$ points on a circle $(n \geqslant 2)$, which are connected by chords, with no three chords intersecting at the same point inside the circle. Let $a_{n}$ denote the number of non-overlapping regions formed inside the circle.
(1) Directly draw to find $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$;
(2) Determine the expression for $a_{n}$.
(1) From Figure 21, we can see that $a_{2}=2, a_{3}=4, a_{4}=8, a_{5}=16, a_{6}=31$.
From the first 4 values, we might guess that $a_{n}=2^{n-1}$, but $a_{6}$ disproves this guess.
(2) Generally, we divide the required area into two parts: one part consists of $n$ segments, and the other part is the region inside the polygon $A_{1} A_{2} \cdots A_{n}$ divided by the diagonals. For convenience, we call these regions inner regions, and the intersection points of the diagonals (denoted as $B_{i}$ in Figure 22) are called nodes.

Consider the case where $n \geqslant 4$ (otherwise, there are no nodes, and the generality is lost). Let the number of triangles in the inner regions of the polygon be $N_{3}$, the number of quadrilaterals be $N_{4}$, ..., and the number of $n$-gons be $N_{n}$. Then the number of inner regions is
$$
m=N_{3}+N_{4}+\cdots+N_{n} \text{ (regions). }
$$

Thus, $a_{n}=m i n$.
The key is to find $m$. We proceed in 3 steps.
(i) First, calculate the expression for the total number of vertices in all inner regions. First,
$$
S=3 N_{3}+4 N_{4}+\cdots+n N_{n} \text{. }
$$

Since each node inside the polygon corresponds to 4 vertices of the polygon (for example, $B_{i}$ corresponds to $A_{1}, A_{2}, A_{3}, A_{n}$ in Figure 22), there are $\mathrm{C}_{n}^{4}$ nodes, and each node corresponds to 4 regions. Each vertex of the polygon can connect to $(n-2)$ diagonals, which are the common points of $(n-2)$ triangles. Therefore, $S$ can also be expressed as
$$
\mathrm{S}=4 \mathrm{C}_{n}^{4}+(n-2) n \text{, }
$$

i.e., $3 N_{3}+4 N_{4}+\cdots+n N_{n}=4 \mathrm{C}_{n}^{4}+n(n-2)$.
(ii) Next, calculate the expression for the total sum of the interior angles of all inner regions. First,
$$
T=180^{\circ} N_{3}+360^{\circ} N_{4}+\cdots+(n-2) 180^{\circ} N_{n} \text{. }
$$

Since each node contains a full angle, the total sum is $360^{\circ} \mathrm{C}_{n}^{4}$. The sum of the angles at each vertex of the $n$-gon is $(n-2) 180^{\circ}$, so we also have
$$
T=360^{\circ} \mathrm{C}_{n}^{4}+(n-2) 180^{\circ} \text{. }
$$

Comparing the two expressions for $T$, we get
$$
N_{3}+2 N_{4}+\cdots+(n-2) N_{n}=2 \mathrm{C}_{n}^{4}+(n-2) \text{. }
$$
(iii) Find $m$, and thus $a_{n}$.

From (1) - (2), we have
$$
\begin{array}{l}
N_{3}+N_{4}+\cdots+N_{n} \\
=\mathrm{C}_{n}^{4}+\frac{(n-2)(n-1)}{2}=\mathrm{C}_{n}^{4}+\mathrm{C}_{n-1}^{2} .
\end{array}
$$

Thus, $a_{n}=\mathrm{C}_{n}^{4}+\mathrm{C}_{n-1}^{2}+n$.
This formula can also be expressed as
$$
\begin{aligned}
a_{n} & =\mathrm{C}_{n}^{4}+\mathrm{C}_{n-1}^{2}+\mathrm{C}_{n}^{1} \\
& =\mathrm{C}_{n-1}^{4}+\mathrm{C}_{n-1}^{3}+\mathrm{C}_{n-1}^{2}+\mathrm{C}_{n-1}^{1}+\mathrm{C}_{n-1}^{0} .
\end{aligned}
$$

If we agree that $\mathrm{C}_{n}^{m}=0(m<n)$, then the general term formula for $a_{n}$ can be any of the above expressions. Since
$$
\begin{aligned}
2^{n-1} & =(1+1)^{n-1} \\
& =\mathrm{C}_{n-1}^{0}+C_{n-1}^{1}+\cdots+\mathrm{C}_{n-1}^{n-1},
\end{aligned}
$$

the first 5 terms of (3) and (4) are the same, and for $n \geqslant 6$, the guess that $a_{n}=2^{n-1}$ is incorrect.