[lanna]

Well, how to calculate sum from 1 to 5000 ?

Instead of looping from 1 to 5000, you compute (1 + 5000) * 5000 / 2

This is what "understand problem first" actually means

[roelschroeven]

But now you can run into problems with integer overflow. It can be solved:

    sum(n) = is_even(n) ? n/2 * (n+1) : (n+1)/2 * n

because of one n and n+1 will always be even. But it show that unfortunately things are often not as straightforward as they first appear.