Example 1. In a plane, there are $\mathrm{n}$ lines $(\mathrm{n} \in \mathrm{N})$, among which no two are parallel, and no three intersect at the same point. Question: How many regions does these $\mathrm{n}$ lines divide the plane into?
Let's assume that $k$ lines divide the plane into $f(k)$ regions. Then, $(k+1)$ lines will divide the plane into $f(k+1)$ regions.

The $(k+1)$-th line $l_{k+1}$ intersects with the original $k$ lines, creating $k$ intersection points. These $k$ points divide the line $l_{k+1}$ into $(k+1)$ segments, and each segment divides its region into two. This leads to the recursive relation:
$$
f(k+1)-f(k)=k+1 \text{. }
$$

For this equation, let $k=1,2,3, \cdots, (n-1)$,
we get: $f(2) - f(1) = 2$,
$f(3) - f(2) = 3$,
$f(4) - f(3) = 4$, a total of $(n-1)$ equations ......
$f(n) - f(n-1) = n$

Adding all these equations, we simplify to:
$$
\begin{aligned}
f(n) - f(1) & = 2 + 3 + 4 + \cdots + n = \frac{(n-1)(n+2)}{2} . \\
\because f(1) & = 2, \\
\therefore f(n) & = \frac{(n-1)(n+2)}{2} + 2 \\
& = \frac{1}{2}\left(n^{2} + n + 2\right) .
\end{aligned}
$$

Thus, these $n$ lines divide the plane into $\frac{1}{2}\left(n^{2} + n + 2\right)$ regions.