For example, through two points $A$ and $B$ on a plane, there are $m$ and $n$ lines respectively. How many parts can these $m+n$ lines divide the plane into at most? $(m n \neq 0)$.
Analysis and Solution This is a binary sequence problem. Let the $m+n$ lines divide the plane into the maximum of $f(m, n)$ parts. It is obvious that $f(1,1)=4$. To find the recursive relationship, we can start with simple cases, first examining the case where $m=1$. At this time, $1+n$ lines can divide the plane into $f(1, n)$ parts, and there is $f(1, n)=f(1, n-1)+3$. This is because when a line is added at point $B$, it is divided into three segments or rays by point $B$ and the line passing through point $A$, increasing the number of existing plane parts by the corresponding number of segments or rays. Iterating the above formula yields
$$
\begin{aligned}
f(1, n) & =f(1, n-1)+3=f(1, n-2)+3 \times 2 \\
& =\cdots=f(1,1)+3(n-1)=3 n+1
\end{aligned}
$$

Similarly, if point $B$ already has $n$ lines, then when a line is added at point $A$, it can have at most $n$ intersection points with these lines, and the added line is divided into $n+2$ segments by these $n$ lines and point $A$. Each segment divides its region into two, i.e., it increases the number of regions by $n+2$, so we have
$$
\begin{aligned}
f(m, n) & =f(m-1, n)+n+2=f(m-2, n)+2(n+2) \\
& =\cdots=f(1, n)+(m-1)(n+2) \\
& =m n+2 m+2 n-1
\end{aligned}
$$