Little Sugar : Big Oh

Consider a scenario where you had a list of N numbers . Now say you have an algorithm that finds the does some sort of comparison . If comparison strategy used compares every element with all the succeeding elements . Then on an average the number of comparisons possible in different cases would be something like this :

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

for the 1 st case we have n comparisons
for the 1 st element we have n-1 comparisons
for the 2 nd element we have n-2 comparison .. and so on.

the sum is = ((n)*(n+1))/2 = O(n^2)