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Calculating sum of a series

  1. Nov 15, 2006 #1
    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


    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. jcsd
  3. Nov 15, 2006 #2
    Is this what you need to find [tex]\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., [itex]\frac{1}{3}\left(r^3 - (r-1)^3 - 1\right)[/itex] and then sum it from 1 to n?
    Last edited: Nov 15, 2006
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