For example, $4 n$ lines intersect each other pairwise, but no three lines intersect at the same point. How many parts will such $n$ lines divide the plane into?
Solution: Let the $k$-th line divide the plane into $a_{k}\left(a_{1}=2\right)$ parts. It is easy to see that the $(k+1)$-th line can intersect the previous $k$ lines at most at $k$ points, and these $k$ points divide the $(k+1)$-th line into $k+1$ segments (including two rays). Each segment divides the region it lies in into 2 parts, i.e., each segment increases the number of regions by 1. Therefore, the $(k+1)$-th line increases the original $a_{k}$ parts by $k+1$. So,
$$
a_{k-1}=a_{k}+k+1(k=1,2, \cdots, n-1) .
$$

Adding these $n-1$ equations, we get
$$
\begin{aligned}
a_{n} & =a_{1}+2+3+\cdots+n \\
& =\frac{1}{2} n(n+1)+1 .
\end{aligned}
$$