a) Let $n$ lines be drawn in the plane so that each of them intersects the others but any three lines do not coincide at any point. For $n\geq 0$, let $a_n$ be the number of regions into which the plane is separated by these $n$ lines. Find and solve a recurrence relation for $a_n$.
b) For the situation in part a), let $b_n$ be the number of infinite regions obtained that way. Find and solve a recurrence relation for $b_n$.
a) If we look at what happens when we add the nth line, it will intersect with all the other $n-1$ lines, so it will pass through $n$ regions, thus dividing them in two. This means that the number of regions will increase by $n$. Therefore, our recurrence relation is:
\begin{align*}
a_n-a_{n-1}=n,\ n>0,\ a_0=1.
\end{align*}
We solve it this way in a simple way:
\begin{align*}
a_n=a_{n-1}+n=a_{n-2}+(n-1)=a_0+1+2+a_{n-2}+(n-1)=a_0+1+2+a_0+1+2+\dots +n.
\end{align*}
Therefore, we have:
\begin{align*}
a_n=\frac{n^2+n+2}{2},n\geq 0.
\end{align*}
b) Let $b_n =$ be the number of infinite regions resulting in $n$ such lines. When the nth line is drawn it is divided into $n$ segments. The first and the nth segment each create a new infinite region. Therefore,
\begin{align*}
b_n=b_{n-1}+2,\ n>1,\ b_1=2
\end{align*}
I don't know if this is correct for a), since for example
By adding the red line, it fulfills the given conditions, but does not divide all regions in two.
Best Answer
Your answer to (a) is correct. This is called the Lazy Caterer's sequence, and is one more than the triangle numbers
Your answer to (b) is also correct for $n\ge 1$ and could be written as $b_n=2n$. In effect there are $n$ lines so $2n$ rays coming out of the area separating $2n$ infinite areas
Your diagram shows $6$ areas ($5$ on the boundary) from $3$ cuts and $10$ ($7$ on the boundary) from $4$ cuts, but you should have $7$ ($6$) and $11$ ($8$). The missing area should be in the top middle, and would appear if you were to move the top middle intersection slightly below the boundary