Into how many parts do $n$ lines in general position divide the plane, that is, such that no two are parallel and no three pass through the same point?
We will prove by induction that the number of parts is $1+1 / 2 n(n+1)$. The base case ($n=1$) is obvious.

Inductive step. Let $n>1$. By the inductive hypothesis, before the $n$-th line is drawn, there were $1+1 / 2(n-1) n$ parts. The new line is divided into $n$ segments by the points of intersection with the old lines. Each of these segments splits one part into two. Therefore, $n$ new parts will be added. Thus, the total number of parts will become $1+1 / 2$ $(n-1) n+n=1+1 / 2 n(n+1)$.

## Answer

$1+1 / 2 n(n+1)$.