sorry isn't a physics question more to do with maths.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Calculating sum of a series

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