So I was doing a random proof from Transition to Advanced Mathematics; it goes as follows:
Let $P_1, P_2,... P_n$ be n points in a plane with no three points collinear. Show that the number of line segments joining all pairs of points is $(n^2 - n) / 2$.
I played around with the problem for a while and I ended up writing the following recurrence: $$F(1) = 0 \\ F(n + 1) = n + F(n)$$
The intuition behind this is that each and every additional point added (not collinear-ly with any other two) simply adds a line segment with each point already present in the plane, which is directly expressed as $F(n)$ which is the amount without the new point and $+\ n$ which is the number of new segments, hence the recurrence. (Note the n + 1 for the second equation.)
Then I noticed that the recurrence relation simply unfolds to: $$F(n) = \sum_{i = 1}^{n - 1}i$$
The intuition behind this one is now rather apparent, as the total number of segments is just the sum of all the segments between points, and we always form $n-1$ new points with each nth point.
So far so good. However, while the term $(n^2 - n) / 2$ is certainly equivalent to the other two I've mentioned (proof by straightforward induction on n, and just rather apparent algebraically), I fail to see how one could intuitively come up with it from the geometrical representation of adding points to a plane. Is there a good way to look at this geometrically, or is the term in this particular scenario just an "artefact"?
Technical note: My question has arisen while looking at this particular proof toy problem, yet I am not asking about the proof itself.
Incrementally from adding point in the plane it's hard to reach the non-recursive term $(n^2-n)/2$. But if you look at all $n$ points in a static setup, you could argue like this:
A line is connecting two points. So take any of the $n$ points as starting point, and any as end point of the line segment. Thus $n^2$ possible pairs. But that includes $n$ pairs where both endpoints are the same; we should remove them. And then our lines are undirected, so the pair $(A,B)$ and the pair $(B,A)$ are the same line. We counted each line twice, so we divide the sum by two.
Somtimes it's easier to think about connecting each point in the plane to each other point. In that case you end up with the equivalent $n(n-1)/2$.