So many lines on one plane??

$\textbf{Recursive solution :}$ .

Let us denote the number of segments divided by $n$ lines to be $L_{n}$ .

Clearly the base case is $L_1 = 2$ .

Now let us assume that $n-1$ lines have already been drawn and we are drawing the $n$ th line.Each time this new line intersects one of the old lines, a new segment is created and also when the line ;eaves the plane it creates an additional segment .Hence we have $L_{n} = L_{n-1} + n$ . $L_{n} = L_1 + (1+2+3+4+....+n)$ $L_{n} = 1 + n(n+1)/2$

So $L_{200} = \boxed{20100}$

This question looks harder than it actually is. I have a rather simple solution.

In order to find the maximum number of segments, you must assume that all the lines intersect each other and that there are only two lines at an intersection. You have to find the value of $1 + 2 + 3 + \ldots + 200$ .

$1 + 2 + \ldots + 200 = S$

$200 + 199 + \ldots + 1 = S$

$\overline{201 + 201 + \ldots + 201 = 2S}$

$2S = 201 \times 200$

$S = 201 \times 100$

$S = 20100$

This means there are $20100$ possible segments that can be formed. But we have forgotten that we started off with $1$ segment, the plane itself. Therefore, the maximum number of segments is $20100 + 1 = 20101$ . :D

I solved this by experiment. The answer follows the formula 0.5(n^2+n+2).

CAN'T WE USE THE SIMPLE FORMULA

$\frac {n^2+n+2}{2}$

And n is 200 so substituting gives 20101

$1^{st}$ Method Creating a rectangle, a simple plane.

Let $f(l)$ be the maximum segments of a plane which can be divided by $l$ lines.

So, by drawing,

Draw one line, the rectangle is divided into two,

$f(1) = 2$

Draw 2 lines, the rectangle will divided into four,

$f(2) = 4$

Draw 3 lines, the rectangle will be maximumly divided into seven.

$f(3) = 7$

So, observing the patterns, we get that

$f(n) = 1 + \sum_{i = 1}^{n}i$

Thus, $f(n) = 1 + \frac{n(n + 1)}{2}$

Therefore, $f(200) = \boxed{20101}$

$2^{nd}$ Method

I recommend you to get an A4 paper before reading this on.

First, draw one line straight across.

We will observe that for every line drewn which intersects another line, one more segment.

So, by this, we need to get our new line to intersect as many lines as possible.

Second, aim for an angle which will intersect most number of lines.

In this case, a new line can only intersect the old line, creating one intersection point, which will give us 4 segments.

Repeat the second step for the third line, the maximum intersections achievable is $2$ , creating 3 segments. This will divide our paper into $7 = 4 + (3)$ segments.

Repat the second step for the fourth line, the maximum intersections achievable is $3$ , creating 4 segments. This will divide our paper into $11 = 4 + (3 + 4)$ segments.

The pattern will go on, thus for the $200^{th}$ line, the maximum intersections achievable is $199$ , creating 200 segments. This will divide our paper into $4 + \sum_{i = 3}^{200}i$ segments.

$4 + \sum_{i = 3}^{200}i$

$= 1 + \sum_{i = 1}^{200}i = \boxed{20101}$

I'll post the solution using sequences instead, everyone just posted the easiest one (for me). XD

Let $x_{n}$ be the max. number of regions divided by $n$ lines.

When you divide, you have to make as much spaces as possible. So you shouldn't be cutting with 3 lines having a common point. When this happens, move 1 of the 3 lines and there'll be more intersections. More intersections = More regions!

Trying to do some drawing and we get that...

$n=1; x_{1} = 2$

$n=2; x_{2} = 4$

$n=3; x_{3} = 7$

$n=4; x_{4} = 11$

After we see the pattern, we see that it jumps every 2,3,4,... times.

We can find the number of regions for $n=200$ .

$x_{200} = 2 + 2 + 3 + 4 + ... + 200 = 1 + \frac{200\times 201}{2} = \boxed{20101}$ ~~~

total no of parts by n lines=(n^2+n+2)/2

There is a simple formula - the maximum number of segments of a plane by n number of lines is given by= 1 + [n(n+1)/2]. substituting n=200 gives 1+ [200(200+1)]/2=1+ 20000 +100=20101