Number of Regions in Plane Defined by Given Number of Lines

Theorem
The maximum number $L_n$ of regions in the plane that can be defined by $n$ straight lines in the plane is:
 * $L_n = \dfrac {n \left({n+1}\right)} 2 + 1$

Setting up a Recurrence Rule
First we consider the plane with no lines at all. This has one region, so $L_0 = 1$.

Now when we have one line, we divide the plane into two regions, so $L_1 = 2$.

Now consider the $n$th line.

This increase the number of regions by $k$ iff it splits $k$ of the old regions.

It can split $k$ of the old regions iff it hits the existing lines on the plane in $k-1$ places.

Two straight lines can intersect in at most one point.

So the new line can intersect the $n-1$ old ones in at most $n-1$ different points.

Therefore $k \le n$.

So we see that $L_n \le L_{n-1} + n$.

Now, it is always possible to place the $n$th line so that:
 * It is not parallel to any of the others, and therefore intersects all the other $n-1$ lines;
 * It does not go through any of the existing intersection points (so intersects them all in different places).

Thus we see that $L_n \ge L_{n-1} + n$.

Hence the recurrence:
 * $L_n = L_{n-1} + n$

Solution of Recurrence
Using induction, we show that $L_n = \dfrac {n \left({n+1}\right)} 2 + 1$.

The base case is straightforward:
 * $L_0 = 1 = \dfrac {0 \left({0+1}\right)} 2 + 1$
 * $L_1 = 2 = \dfrac {1 \left({1+1}\right)} 2 + 1$

Now assume the induction hypothesis:
 * $L_k = \dfrac {k \left({k+1}\right)} 2 + 1$

and try to show:
 * $L_{k+1} = \dfrac {\left({k+1}\right) \left({k+2}\right)} 2 + 1$

Hence the induction step:

Hence the result by induction.

History
This was shown by Jakob Steiner in 1826 in Einige Gesetze über die Theilungder Ebene und des Raumes, published in Journal für die reine und angewandte Mathematik 1: 349-364.