The size of the number (result) is (N(N + 1))/2, the computation of the algorithm is constant. I've never heard anyone discuss the result of the algorithm using Big O.

I think he meant factorial rather than sum.

The sum of the numbers is O(1) - you can calculate it algorithmically.

N*(N+1)/2

Ref: http://www.wikihow.com/Sum-the-Integers-from-1-to-N