Triangular Combinations

          During my work on polynomial expansions, I came across this pleasing result.  Enjoy!

                    ²C₂ = 1

                            = (2 - 1)

                    Assuming that ^kC₂ = 1 + 2 + ... + (k - 1),

                    To prove that ^{k + 1}C₂ = 1 + 2 + ... + k,

                    Note that to choose 2 from a set with 1 more item than the previous, we simply include the new item and find the number of combinations of two that include that item.  The new sets contain the number in the old set, each combined with the new item, and hence we simply add k.  So:

                     ^{k + 1}C₂ = ^kC₂ + k

                                   = 1 + 2 + ... + (k - 1) + k    (as required)

                     ∴ if true for k, it is true for k + 1

                     But it is true for 2, so it is true for 2 + 1 = 3, 3 + 1 = 4 and so on for all integers > 2

                     ∴ By induction, all combinations of 2, in a set of x, are the triangular number for n = x - 1