(thing) by mblase Wed Jan 17 2001 at 16:37:37

Think of a rack of billiard balls, or Pascal's Triangle. You start with 1, add two to get 3, add three to get 6, add four to get 10, add five to get 15, and so forth, each time adding the next natural number to the previous sum. The numbers you produce using this algorithm are the triangular numbers. 

        1
      2   3
    4   5   6
  7   8   9  10
11 12  13  14  15

The nth triangular number is defined as the number of elements in a triangle with n rows, that is: n+(n-1)+(n-2)+...+2+1.

Because of this, the nth triangular number can be quickly calculated for any even n:

The nth triangular number
= n+(n-1)+(n-2)+...+(n-n/2)+...+2+1
= n+(n-1)+1+(n-2)+2+...+(n-n/2)
= n+n+n+...+n (repeated n/2 times) +(n-n/2)
= n*(n/2)+n-(n/2)
= n(n/2+1-1/2)
= n(n/2+1/2)
= n(n+1)/2
Similarly, for any odd n:
The nth triangular number
= n+(n-1)+(n-2)+...+(n-(n-1)/2)+(n-1)/2+...+2+1
= n+(n-1)+1+(n-2)+2+...+(n-((n-1)/2)+((n-1)/2))
= n+n+n+...+n (repeated 1+(n-1)/2 times)
= n+n+n+...+n (repeated (n+1)/2 times)
= n(n+1)/2
We can generalize that the nth triangular number equals n(n+1)/2 for any natural number n, and if we like, prove it using mathematical induction.

I like it!