A geometric question about drawing lines in a plane Suppose we were to draw lines in a plane such that no two of them are parallel and no three or more meet at one point, so in other words there is only double intersections. If we drew $x$ lines how many regions would be bounded and how many regions would be unbounded? So for example, with $3$ lines we would have $6$ unbounded regions and $1$ bounded region. 
My idea was as follows. Find a pattern pertaining to how many lines make how many regions then write an equation to represent this pattern and proceed with induction. Problem is with the table I have, I dont see a particular pattern yet. $0$ lines gives us $1$ region, $1$ line gives us $2$ regions, $2$ lines gives us $4$ regions, $3$ lines gives us $7$ regions, $4$ lines gives us $11$ regions, etc. 
 A: For starters, I must have that there are 2 parts of the plane for every line I draw, ignoring all intersections. Great, but this doesn't help us entirely, because we still have do deal with what happens where lines intersect. Now consider that every line intersects with every line once as you draw it (since it's not parallel to any others, this is axiomatic in geometry). Every time we intersect with one line and then another we get a new region of the plane, and whenever we come in from "outside" of our tangle of intersections, as I'll call it, or go back "outside" again, we also get a new partition.
In particular, consider the following: we start with one line. 2 sections of the plane. This is our base case. Then we draw another. Easy, since there's only one possible situation. You draw your line, coming 'into your tangle' which gives you a new third section by splitting one of your first two, and you then draw the rest of it 'going out', giving you number 4.  Now consider what happens at the $n$th step: you go in (+1 section), and then intersect with all $n-1$ lines that were already there (+n-2 sections), and then you leave the tangle again (+1 section) and you'll never intersect with anything else again. Shockingly, there's still exactly one possible situation, just like cases 1 and 2!
So at the $n$th step, you get $n$ new sections of the plane. This means that to draw $n$ lines in the plane is to cut it into $\left(\sum_{i=1}^n i\right)+1$ pieces. Count that up and you're done.
Next, you want to know about inside and outside regions. Well, we also set up our story to deal with that case. With each line drawn you get $n-1$ new inside regions and 2 outside ones, and so the number of inside regions is going to be $\left(\sum_{i=1}^n i-2\right)+1$, while the number of outside regions is going to be $2n$.
A: Try searching the Online Encyclopedia of Integer Sequences and you'll get referred to sequence A000124, the "Central polygonal numbers (the Lazy Caterer's sequence): n(n+1)/2 + 1; or, maximal number of pieces formed when slicing a pancake with n cuts."
As for the number of bounded and unbounded regions, you can first get the few cases then search the OEIS again to see which sequence best fits your situation.
