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Summing Series (Sigma Notation)

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data

    a) nƩr2(r-1)r=1

    2. Relevant equations

    Using the summation series formulae...

    3. The attempt at a solution

    So far I have got:

    r2(r-1) = r3-r2

    Ʃr3 = [itex]\frac{1}{4}[/itex]n2 (n+1)2

    Ʃr2 = [itex]\frac{1}{6}[/itex]n(n+1)(2n+1)


    Ʃr3-r2 = [itex]\frac{1}{4}[/itex]n2 (n+1)2 - [itex]\frac{1}{6}[/itex]n(n+1)(2n+1)

    But how do I subtract these two massive things from each other?!?!

  2. jcsd
  3. Nov 22, 2011 #2


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    Homework Helper

    They aren't that massive. Rewrite them as fractions:
    Find a common denominator, rewrite both fractions so that both have that common denominator, and then subtract. It will be possible to factor things out.
  4. Nov 22, 2011 #3


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    Science Advisor

    It's pretty much just subtracting fractions. Get a common denominator:
    The least common denominator of 4 and 6 is 12. To change the first denominator to 12, multiply both numerator and denominator by 3.
    [tex]\frac{n^2(n+1)^2}{4}= \frac{3n^2(n+1)^2}{12}[/tex]

    To change the second denominator to 12, multiply both numerator and denominator by 4.

    To subtract now, subtract the numerators:
    [tex]\frac{3n^2(n+1)^2}{12}- \frac{4n(n+1)(2n+1)}{12}[/tex]

    You will need to multiply those:
    [itex]3n^2(n+1)^2= 3n^2(n^2+ 2n+ 1)= 3n^4+ 6n^3+ 3n^2[/itex]
    [itex]4n(n+1)(2n+1)= 4n(2n^2+ 3n+ 1)= 8n^3+ 12n^2+ 4n[/itex]

    and now subtract.
  5. Nov 22, 2011 #4
    Yeah ok, I've done that, but the textbook answers want it to be "factorised", and I'm unsure how to do that after subtracting... Thanks for your help.
  6. Nov 22, 2011 #5

    Hmm thanks, for the help, I'll try to factorise those now, since the book wants it to be factorised.
  7. Nov 23, 2011 #6


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    Science Advisor

    You will have a quadratic factor that cannot be factored further (with integer coefficients).
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