I was wondering a formula F(n) to guess the maximum number of faces made with n lines, for example:
with 1 line, we cant create a face;
F(n) = 0;
with 2 lines, we also cant create a face
F(n) = 0;
with 3 lines, the best we can do is a triangle;
F(n) = 1;
with 4 lines, the best we can do is split that triangle in 2 and cross another line, making 3 faces...
F(n) = 3;
Searching about that, I found the Lazy caterer's sequence, that given a number n of lines, it divides a 2D space in at most $\frac{ n² + n + 2}{2}$ parts, but some of this parts are not faces, they are just a space created by lines that do not close (for example, there are 4 of this spaces if you cross 2 lines, and they are not faces).
So I thought:
if the lines go to the infinity, all n lines together will create 2n of this spaces that are not faces
[suppose you look at them from a distance so far that the part where occurs the intersections seems to be a point, then all the lines would be crossing a single point and then, creating 2n spaces between them]
if there are n spaces, then the number of faces would be:
$$ F(n) = \frac {n^2 + n + 2}{2} - 2n $$ that is
$$ F(n) = \frac {n^2 - 3n + 2}{2} $$
Am I right? If so, how can I prove that? I'd like to try with induction, but I dont know how...