Question need to get a conjecture and prove by induction 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?
 A: Conjecture: the number of regions = [(2+n)(n-1)]/2 +2 = (n^2 +n +2)/2
A: 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.
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.
