Example 3: On a plane, there are $n$ lines, no two of which are parallel, and no three of which intersect at the same point. How many intersection points do these $n$ lines have in total?
Example 4: Following the previous example, now we want to know how many non-overlapping regions will be formed by such $n$ lines dividing the plane?
Solution: Let the number of intersection points that meet the requirements for $n-1$ lines be $a_{n-1}$. When adding the $n$-th line, the new line intersects with each of the previous $n-1$ lines at exactly one point. Therefore, the number of intersection points that meet the requirements for $n$ lines is $a_{n}=a_{n-1}+(n-1), a_{2}=1$. Solving this gives $a_{n}=$ $\frac{n(n-1)}{2}$

Solution: Let the number of regions formed by $n-1$ lines be $a_{n-1}$, then $a_{1}=2, a_{2}=4$. When adding the $n$-th line, it is equivalent to adding $n-1$ points on the $n$-th line, dividing this line into $n$ segments (2 of which are rays), and each segment divides the original region into two, thus adding $n$ regions. Therefore, the number of regions formed by $n$ lines is $a_{n}=a_{n-1}+n$, solving this gives $a_{n}=$ $\frac{n(n+1)}{2}+1$