sorry isn't a physics question more to do with maths. is it ok to post it here? I can't see to delete it, maybe some one will move it for me? to the maths section. thanksL:) I have to calculate: sum of (from r=1 to n): 1/3 (r^3 - (r-1)^3 -1) so far i've done: let n = 3 so sum = 1/3 (n^3 - (n-1)^3 -1) +1/3 ( (n-1)^3 - (n-2)^3 - 1) +1/3 ( (n-2)^3 - (n-3)^3 -1) looking at ^ (n-3)^3 = 0 the 1/3 can be factorised out, the -1's sums to -n so: sum = 1/3 (n^3 -...........-n) (1) this inbetween thing im finding tricky i noticed theres....-2(n-1)^3 - 2(n-2)^3...etc, when n = any number. i can put that as sum(from r=1 to n-1) of: (n-r)^3 so into (1): 1/3 (n^3 -2(sum from r=1 to n-1 ofn-r)^3) -n) ^but i don't think that's what they are looking for. any help will be appriciated,thank you.