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Sum of a series

  1. Jan 21, 2007 #1
    1. The problem statement, all variables and given/known data
    Find a general formula for
    [tex] (1\cdot 2) +(2\cdot 3) +... + (n\cdot (n+1)) [/tex]

    2. Relevant equations
    The sum of the first n integers
    [tex] 1+2+3+... + n = \frac{n(n+1)}{2} [/tex]

    3. The attempt at a solution
    i thought of doing this
    [tex] (1+2+3...) * (2+3+4+...) [/tex]

    but then the problem of cross terms comes about

    additionally this is not an arithmetic series nor a geometric series because the common difference increases by 1 each term and there is no common factor.

    how wouldi get rido of the cross terms??
    do i even need to do it this way?? Another method is possible??

    Please help!!

  2. jcsd
  3. Jan 21, 2007 #2


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    Do you know the formula for [tex]\sum_{i=1}^n i^2[/tex]?
  4. Jan 22, 2007 #3
    So you want to sum:

    1*2 + 2*3 + 3*4 + 4*5 + ... + n(n+1)

    It's very difficult working with numbers. You'll never get anywhere. So the first order of business is to rewrite the above sum using a symbolic sum:

    1*2 + 2*3 + 3*4 + 4*5 + ... + n(n+1) = \sum_{i=start}^{end} foo

    You shouldn't have too much difficulty figuring out what "start", "end" and "foo" should be to represent the left side of that equation.

    You'll see that "foo" is a product of two terms. They'll look "factored". Unfactor them (distribute terms) and you'll end up with a sum of two terms. It'll look like:

    \sum_{i=start}^{end} hello + kitty

    One term is Quasar's term. The other term you'll want to ask Mr. Gauss about.

    Note: You can sum some of the series some of the time, but you can't sum all of the series all of the time. This case happens to be one of the "some of the time".
    Last edited by a moderator: Jan 22, 2007
  5. Jan 22, 2007 #4
    so then the sum of the series is

    [tex] \sum_{i=1}^{n} i(i+1) = \sum_{i=1}^{n} i^2 + \sum_{i=1}^{n} i [/tex]


    thanks guys
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