Discrete mathematics proof that I have been stuck on So I have been working on these proofs for a while and finished 13 of 14 of them but I was never able to figure this one out so I thought I would ask for help on how it would be done:S
Here is the question:
Let $n\ge$ be an integer. Consider $2n$ straight lines $L_1,L_1',L_2,L_2',\ldots,L_n,L_n'$ such that:


*

*For each $i$ with $1 \le i \le n$, $L_i$ and $L_i'$ are parallel.

*No two of these lines $L_1,\ldots,L_n$ are parellel,

*No two of these lines $L_1',\ldots,L_n'$ are parallel

*No three of the $2n$ lines intersect in one single point.


These lines divide the plane into regions (some of which are bounded and some of which are unbounded). Denote the number of the regions by $R_n$. 
Derive a recureence from the number $R_n$ and use it to prove that $R_n = 2n^2 + 1$ for $n\ge1$
I don't understand how I should be going about this problem, I think the question is unclear (atleast in my mind, hopefully someone can give me the general direction I could be going with this though). Any help at all would be tons of help. Thanks!
 A: Say you have $r_n$ regions when there are $2n$ lines. Now you add $L_{n+1}$; it’s not parallel to any of $L_1,\ldots,L_n$ or $L_1',\ldots,L_n'$, so it eventually crosses every one of those $2n$ lines. That means that it runs through $2n+1$ of the regions formed by the first $2n$ lines. (Why?) Whenever it runs through a region, it splits that region in two. By how much does this increase the number of regions? Now add $L_{n+1}'$ parallel to $L_{n+1}$. How many regions does it cut in two? By how much does that increase the number of regions?
Once you have all that, you can express $r_{n+1}$ in terms of $r_n$ and $n$ to get your recurrence.
A: As you draw each line, count how many other lines you cross. Relate the number of other lines you cross to how many regions you add.
A: This is a nonrecursive way using the Euler formula $V-E+F=1$ where $V,E,F$ are the numbers of vertices, edges, and faces of the final graph. We first need to find $V,$ the number of vertices. Each of the $2n$ lines intersects all but its parallel companion line, so it meets $2n-2$ other lines. The expression $2n(2n-2)$ would "double count" the vertices, since each vertex is on two of the lines, so we have $V=n(2n-2).$
We also need $E,$ the number of edges. Each line has $2n-2$ vertices on it, so these divide that line into $2n-1$ edges of the graph. So we have $E=2n(2n-1).$
Finally, solving the Euler equation for $F$ gives the result
$$F=E+1-V=2n(2n-1)+1-n(2n-2)=2n^2+1$$
for the number of faces or "regions".
Note: I know the OP specifically requested a recurrence, but only include this as an alternate way to look at the problem.
