This is mathematical proof, but I guess this is fairly intuitive too..

Line up the series of numbers from 1 to (n-1) once, and in reverse right below that:

1 + 2 + 3 + ...................+ (n-2) + (n-1)

(n-1) + (n-2) + (n-3) +..................... + 2 + 1

Adding these vertically, these are (n-1) pairs of n.... So the vertical sum is n*(n-1), and since this is twice the sum that we originally wanted.

So, required sum is n*(n-1)/2 or n choose 2.

See question on Quora