1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    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. 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)
    [/tex]?

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...