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Draw $n$ lines on the plane in such a way that no two are parallel and no three intersect in a common point. Prove that the plane is divided into $$\frac{n(n+1)}{2}+1$$ regions. Is we prove using mathematical induction.

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closed as off-topic by Math1000, user133281, iadvd, suomynonA, user91500 Oct 7 '16 at 6:29

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  • $\begingroup$ Hint: Does the expression $\frac{n(n+1)}2$ look familiar? $\endgroup$ – Math1000 Oct 6 '16 at 9:06
  • $\begingroup$ @Math1000...sorry i was edited now $\endgroup$ – Rajendra Oct 6 '16 at 9:07
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Yes, you can use mathematical induction to prove this result. Obviously for $n=1$ the claim is true. Now assume that it's true for some positive integer $k$.

Take any configuration of $k$ lines such that the conditions are satisfied. Now we know that by inductive hypothesis that there are $\frac{k(k+1)}{2} + 1$ regions. Now add any line into the configuration. Note that the new lines must cut the other $k$ lines in distinct $k$ points, as it's not parallel to any of it and no three lines intersect at one point. Therefore we'll get a new $k$ intersection points. But note that at such an intersection point the new line "stops cutting" the previous region and "starts cutting" the next region. And these are the only such points. Hence we can conclude that by adding a new line we've created $k+1$ new regions. Therefore there are number of regions now is:

$$\frac{k(k+1)}{2} + 1 + (k+1) = \frac{(k+2)(k+1)}{2} + 1$$

Hence the proof.

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