# Homework Help: Calculating sum of a series

1. Nov 15, 2006

### alias25

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.

Last edited: Nov 15, 2006
2. Nov 15, 2006

### neutrino

Is this what you need to find $$\sum_{r=1}^{n}\left(\frac{1}{3}\left({r^3 - (r-1)^3 - 1}\right)\right)$$?

If so, why not simplify the term that is to be summed, i.e., $\frac{1}{3}\left(r^3 - (r-1)^3 - 1\right)$ and then sum it from 1 to n?

Last edited: Nov 15, 2006