Problem source

A plane contains $n$ distinct given lines, no two of which are parallel, and no three of which intersect at a point. By first considering the cases $n=1,2,3$ and $4$, provide and justify, by induction or otherwise, a formula for the number of line segments (including the infinite segments). 
Prove also that the plane is divided into $\frac{1}{2}(n^{2}+n+2)$ regions (including those extending to infinity).

Solution source

With $n=1$ line, the plane is divided in half. 
With $n=2$ lines the plane is divided into four pieces. (Each of the previous pieces are split in half)
With $n=3$ lines the plane is divided into up to $7$ pieces. (The new line crosses two lines in two places dividing $3$ regions into $2$, thus increasing the number of regions by $3$).
With $n=4$ lines the plane is divided into $11$ pieces. (The new line crosses three lines in three places doubling the number of regions of $4$ places).

Claim: With $n$ lines the plane is divided into $\frac12(n^2+n+2)$ regions.

Proof: (By induction) 

(Base case) When $n=1$ clearly the line is divided into $2$ regions, and $\frac12 (1^2 + 1^2 + 2) = 2$ so the base case is true.

(Inductive step) Suppose our formula is true for $n=k$, so we have placed $k$ lines in general position and divided the plane into $\frac12(k^2+k+2)$ regions. When we place a new line it will meet those $k$ lines in $k$ places (since no lines are parallel) and there will be k+1 regions the line will run through (since no three lines meet at a point). Each of those $k+1$ regios is now split in half, so there are $k+1$ "new regions".

Therefore there are now $\frac12(k^2+k+2)+(k+1) = \frac12(k^2+k+1+2k+2) = \frac12 ((k+1)^2+(k+1)+1)$ regions, ie our hypothesis is true for $n=k+1$.

(Conclusion) Therefore since our statement is true for $n=1$ and since if it is true for some $n=k$ it is true for $n=k+1$ by the principle of mathematical induction it is true for all $n \geq 1$

Proof: (Alternative).

There are $\binom{n}{2}$ places where the lines meet. Each intersection is the most extreme point (say lowest) for one region (if two are tied, rotate by a very small amount) so this is a unique mapping. There will be $n+1$ regions which are infinite and don't have a most extreme point, hence $\binom{n}{2} + n+1 = \frac12(n^2-n)+n+1 = \frac12(n^2+n+2)$