10. Derivation
• Two lines cut in 1 point.
• A third line will cut the other two lines in 2 more points, giving 1 + 2 = 3
points.
• A fourth line will cut the other 3 lines in 3 more points, giving 3 + 3 = 6
points.
• So the series goes n = 1, 2, 3, 4, 5, 6, .......Number of points 0,
1, 3, 6, 10, 15, .......The gap increases by 1 each time.
• This is the sequence of triangular numbers which has the nth term given by
n(n-1)/2.
OR
• If you made up a difference table, the second differences would be constant,
so the nth term is a quadratic in n.
• If you assume f(n) = an^2 + bn + c, where f(n) is the nth term n = 1 a +
b + c = 0 n = 2 4a + 2b + c = 1 n = 3 9a + 3b + c = 3 and solving these
3 equations for a, b, c, a = 1/2, b = -1/2, c = 0
• So f (n) = n^2/2 - n/2 = n(n-1)/2 as given above. So with n lines there are n
(n-1)/2 intersection points.