Trying to understand formula for counting regions of hyperplane arrangements in $\mathbb{R}^2$ There are up to $\binom{n}{2} + n + 1$ regions created by a hyperplane arrangement in $\mathbb{R}^2$ containing $n$ hyperplanes.
I want to understand this in a demonstrative way. Each hyperplane splits each region it passes in two regions. A newly added hyperplane therefore creates a new region each time it crosses another hyplerplane or the plane. So the number of regions should be $1$ (plane) $+ \binom{n}{2}$ (number of intersections of $n$ lines). But as I know the answer, I know that I am missing $n$ regions. 
Do you know where my mistake is? Why do I have to add the number of lines to get the max. number of regions? 
 A: Euler characteristic
You are in $\mathbb R^2$, so your “hyperplanes” are actually lines. You can do the counting using e.g. the Euler characteristic:
$$\chi=V-E+F$$
In general position, each of your $n$ lines intersects $n-1$ others, so it is split into $n$ edges. Therefore you get $E=n^2$. You also get $V=\binom n2$ points of pairwise intersection. Together with the Euler characteristic of the plane, which is $\chi=1$, you get the number of regions as
\begin{multline*}
F=\chi-V+E=1-\binom n2+n^2=1-\left(\tfrac12n^2-\tfrac12n\right)+n^2 \\
=\tfrac12n^2+\tfrac12n+1=\left(\tfrac12n^2-\tfrac12n\right)+n+1=\binom n2+n+1
\end{multline*}
Induction
If you prefer, you can also deduce your result inductively. The $n$th line will intersect $n-1$ others, so it is split into $n$ segments (or edges as we already used before). Each of these segments divides an existing region into two. So you get the recursion
$$f(n) = f(n-1) + n$$
Combine that with $f(0)=1$ and you will find that the results are uniquely specified, and your formula can be used to describe them.
Generalizations
The above considerations assume lines in general position, i.e. no three lines intersect in a single point. For a more general investigation which covers these cases as well, see Intersection of lines on a plane.
For generalizations to higher dimensions, you can alsways consider the $n$ segments per line as $n$ regions on a one-dimensional hyperplane arrangement of $n-1$ hyperplanes. In other words, if you have $n$ hyperplanes in dimension $d$, then the last of these will intersect all the others in a pattern which is an arrangement of $n-1$ hyperplanes in dimension $d-1$, and each region of that arrangement will split a region of the $d$-dimensional arrangement.
The current answer by Olivier appears to be reasoning a lot along these lines of general dimensions. And he's using inequalities since the equalities only hold in the general position case mentioned above. So his answer made me think of this last section (thanks a lot for that!), although the formulation is my own.
A: Suppose there are $p$ regions in the arrangement where you only consider the $n-1$ first lines. When adding the $n$-th line $\ell$, you keep all the old regions, and you split certain regions in two. To be precise, you add one extra region for every region in the hyperplan arrangement induced on $\ell$ by the first $n-1$ hyperplanes. This is a hyperplane arrangement in a one dimensional space, so it consists of a finite number ($\leq n-1$) of points in the line $\ell$. There are at most $n-1+1$ regions in this arrangement, so the number of regions in the $n$ line arrangement is at most $p+n$. If we let $M_n$ stand for the maximal number of regions in a plane hyperplane arrangement with $n$ hyperplanes, then $M_1=2$ and for all $n\geq 2$
$$M_{n}\leq M_{n-1}+n$$
Therefore $M_n\leq M_1+2+3+\cdots+n=1+\frac{n(n+1)}{2}=1+n+\binom{n}{2}$.
