Handshakes and triangular numbers
If everyone at a party of n people shakes hands with everyone else exactly once,
how many handshakes are there?
Each person shakes n − 1 hands, which gives n(n − 1), but every handshake was
counted twice, once from each side. So the answer is n(n − 1)/2.
n: 2 3 4 5 6 7 8 handshakes: 1 3 6 10 15 21 28
These are the triangular numbers, the same ones you get by stacking rows of 1, 2, 3, … dots. That is not a coincidence: the newest guest shakes hands with everyone already there, adding one more row.
The young Gauss story
The classroom version asks for 1 + 2 + … + 100. Pair the first and last terms, the second and
second-to-last, and so on: fifty pairs that each add to 101, so 5,050. Whether or not the story about Gauss is
true, the pairing is the cleanest proof of the formula.