Example 3.6 There are $n$ lines on a plane, with no two lines parallel and no three lines concurrent. Find how many disconnected regions the plane is divided into by these $n$ lines.
Solution: Let the plane be divided into $a_{n}$ disconnected regions by these $n$ lines. It is evident that $a_{0}=$ $1, a_{1}=2$. When $n \geqslant 2$, remove one line $l$ from these $n$ lines, then the remaining $n-1$ lines divide the plane into $a_{n-1}$ disconnected regions. Now, return $l$ to its original position. By the problem's condition, $l$ intersects with the other $n-1$ lines and the $n-1$ intersection points are distinct (otherwise, there would be three lines intersecting at one point). These $n-1$ intersection points divide $l$ into $n$ segments, each segment dividing one of the original regions into two smaller regions. Therefore, after returning $l$, the number of disconnected regions equals the sum of the original number of disconnected regions and $n$, so
$$
\begin{aligned}
a_{n} & =a_{n-1}+n \\
& =a_{n-2}+(n-1)+n \\
& =\cdots \\
& =a_{1}+2+3+\cdots+n \\
& =2+2+3+\cdots+n,
\end{aligned}
$$

i.e.,
$$
a_{n}=1+\frac{n(n+1)}{2} \quad(n \geqslant 2) .
$$

It is evident that the above formula still holds when $n=0,1$, so
$$
a_{n}=1+\frac{n(n+1)}{2} \quad(n \geqslant 0) .
$$