3  [Systems of segments, lines, and circles]

How many parts do $n$ lines in general position divide the plane into, that is, such that no two are parallel and no three pass through the same point?

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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

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