1+2+3+...+n is always less than n+n+n...+n n times. you can rewrite this n+n+..+n as n*n.
f(n) = O(g(n)) if there exists a positive integer n0 and a positive
constant c, such that f(n) ≤ c * g(n) ∀ n ≥ n0
since Big-Oh represents the upper bound of the function, where the function f(n) is the sum of natural numbers up to n.
now, talking about time complexity, for small numbers, the addition should be of a constant amount of work. but the size of n could be humongous; you can't deny that probability.
adding integers can take linear amount of time when n is really large.. So you can say that addition is O(n) operation and you're adding n items. so that alone would make it O(n^2). of course, it will not always take n^2 time, but it's the worst-case when n is really large. (upper bound, remember?)
Now, let's say you directly try to achieve it using n(n+1)/2. Just one multiplication and one division, this should be a constant operation, no?
No.
using a natural size metric of number of digits, the time complexity of multiplying two n-digit numbers using long multiplication is Θ(n^2). When implemented in software, long multiplication algorithms must deal with overflow during additions, which can be expensive. Wikipedia
That again leaves us to O(n^2).