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A family of $n$ lines is drawn in the plane such that

  1. each pair of lines cross and

  2. no $3$ dinstinct lines have a point in common

Let $r(n)$ denote the number of regions into which the plane is partitioned by these lines. Evidently, $r(1) = 2$, $r(2) = 4$, $r(3) = 7$ and $r(4) = 11$.

To determine r(n) for all positive integers, it is enough to note that $$\begin{cases}r(1) = 2\\ r(n) = n+r(n-1)\text{ if }n>1\end{cases}$$.

I kind of get the concept of the problem, but I don't get how they came up with this recursive formula. Could someone explain this to me? Thanks!

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    $\begingroup$ See if this explanation helps. $\endgroup$ – Brian M. Scott May 25 '15 at 14:02
  • $\begingroup$ Draw a straight line on a sheet of paper. Count the regions. Then draw another straight line on the same sheet of paper. Count the regions and note how many MORE regions there are than before. Then draw another straight line on the paper and count how many MORE regions you have that were not there before.... $\endgroup$ – Geoffrey Critzer May 25 '15 at 16:37

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