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.

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