[ColinWright]

Naively:

   There are n starting points
   There are n ending points
   The interval is n/2 in length on average
   Thus it requires n * n * (n/2) additions

You said:

    Generating one partial sum costs n, and
    you need to do it n times, ...

I think you need to do it n^2 times.

[rielfowler]

There are n^2 results, right?

[ColinWright]

Yes, and naively each of those results takes O(n) to compute.

[rielfowler]

Let me correct my self, there are n^2 results in total, when the algorithm ends. The naive algorithm is doing n times more work. That is redundant, because there is clearly an overlap of some partial sums.

[ColinWright]

OK, now I don't understand what you are trying to say.

There are n^2 results, each result takes O(n) to compute, so the naive algorithm is O(n^3). That's what you asked - why does the naive algorithm take O(n^3) - and that seems to answer your question.

Yes, clearly there is an overlap of some partial sums, and that's why a less naive algorithm is sub-O(n^3).

So, what are you asking, or what additional point are you making?