help solving for f(n) =

Question

number of points   1    2    3    4   5  N     35

segments                0   1    3    4

if you place 35 points on a piece of paper so that no three points are collinear how many segments are necessary to connect each point to all others

Answer 1

Suppose we have 5 points A, B, C, D, E then the segments can be labelled by their endpoints:

AB, AC, AD, AE, BC, BD, BE, CD, CE, DE (10 segments).

For 35 points P₁, P₂, ..., P₃₅ the following connections apply:

P₁P₂, P₁P₃, ..., P₁P₃₅ (34 segments)

P₂P₃, P₂P₄, ..., P₂P₃₅ (33 segments)

...

P₃₄P₃₅ (1 segment).

So the total number of segments is 34+33+...+1.

These can be grouped in pairs (34+1)+(33+2)+(32+3)+...+(19+16)+(18+17)=17×35=595 segments.

The formula for n such points is f(n)=½n(n-1).

n	f(n)
1	0
2	1
3	3
4	6
5	10