Show that $n$ lines separate the plane into $\frac{n^2+n+2}{2}$ regions 
Show that $n$ lines separate the plane into  $\frac{n^2+n+2}{2}$ regions if no two of these lines are parallel and no three pass through a common point.

I know we start with the base case, where, if we call the above equation P(n), P(0), for 0 lines would be 0.  But I really have no idea how to begin the inductive step.  How do we know what k+1 we're supposed to arrive at?
Thanks!
 A: Suppose we draw $n$ straight lines on the plane so that every pair of
lines intersects (but no $3$ lines intersect at a common point). Into 
how many regions do these $n$ lines divide the plane?
With $n = 1$ we divide the plane into $2$ regions. With $n = 2$ we have $4$ regions; with $n = 3$ we get $7$ regions. A fourth line will meet the 
other $3$ lines in $3$ points and so traverse $4$ regions, dividing them 
into $2$ parts and adding $4$ new regions. In general the $n^{th}$ line will 
add $n$ new regions:
$$u(1) = 2$$
$$u(2) = 4$$
$$u(3) = 7$$
$$u(4) = 11$$ 
And so on, where $u(n) =$ number of regions with $n$ lines.
We get the recurrence relationship:
$$u(n+1) = u(n) + (n+1)$$
We get the following chain of equations:
$$u(n) - u(n-1) = n$$
$$u(n-1) - u(n-2) = n-1$$
$$u(n-2) - u(n-3) = n-2$$
$$\vdots$$
$$u(4) - u(3)   =  4$$
$$u(3) - u(2)   =  3$$
$$u(2) - u(1)   =  2$$
Adding these equations, we get:
$$u(n) - u(1)   =  2 + 3 + 4 + ..... + (n-1) + n$$
All other terms on the left cancel between rows, and we are left with:
$$u(n) = u(1) + 2 + 3 + 4 + \ldots + n$$
We know, $u(1) = 2$
Thus:
$$u(n) = 1 + (1+2+3+4+ \ldots+n)$$
$$\implies u(n) = 1 + \dfrac{n(n+1)}{2}$$
$$\implies u(n) = \dfrac{n^2 + n + 2}{2}$$
So:
$$u(n) = \dfrac{n^2 + n + 2}{2}$$
Remark $\,$ If you allow parallel lines and more than $2$ lines to intersect at a 
point, the above relation doesn't hold.
The answer then depends on the number of lines intersecting at a point or the number of lines which are parallel to one another.
A: Here is the way I usually think about this (it sort of uses induction).
With $0$ lines, there is $1$ region and no intersections of lines.
Each time a line is added and it crosses $k$ other lines it adds $k+1$ regions and $k$ intersections. Another way of looking at this is that for each line and $k$ intersections added, $k+1$ regions are added (the number of added lines and intersections).
Therefore, the number of regions is $1+\text{the number of lines}+\text{the number of intersections}$. With $n$ lines, there are $\binom{n}{2}$ intersections (if no two lines are parallel and no three lines are coincident).
Thus, the number of regions is $\binom{n}{2}+n+1=\frac{n(n-1)}{2}+n+1=\frac{n^2+n+2}{2}$.
