Question need to get a conjecture and prove by induction

Question

The question is:

Divide the plane into separate regions using $N$ lines according to the following rules:

• No two lines are parallel.
• No three lines intersect at the same point.

When $N = 1$, the plane is divided into 2 regions. When two lines are drawn in this way, the plane is divided into 4 regions.

I count there are 2 regions for 1 line, 4 regions for 2 lines, 7 regions with 3 lines, 11 regions with 4 lines, 16 regions with 5 lines $\ldots$

I didn’t get a clue for the conjecture. Please give me a hint to get a conjecture?

Answer 1

Conjecture: the number of regions = [(2+n)(n-1)]/2 +2 = (n^2 +n +2)/2

Answer 2

Let $u_n$ denote the number of regions with $n$ lines so $u_1=2$. When $n$ increasing from $k$ to $k+1$, the new line can intersect each of the original $k$ lines, thereby adding $k+1$ regions, so $u_{k+1}=u_k+k+1$. Any conjectured $u_n$ should be quadratic in $n$. If you use $u_n=an^2+bn+c$ to get simultaneous equations from $u_1=2,\,u_2=4,\,u_3=7$, you can then use the recursion formula in an inductive proof's inductive step.

Answer 3

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.