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Math Help - Trying to find the sum of a series

  1. #1
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    Question Trying to find the sum of a series

    Hello I have found a pretty complicated way of finding the sum to this series, and was wondering if anyone could help me to find a simpler way.

    The sum of the series 1+2+3+4+....+n= n(n+1)/2

    What is the sum of 2+6+20+....+n(n+1)=???
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  2. #2
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    Quote Originally Posted by Finney View Post
    Hello I have found a pretty complicated way of finding the sum to this series, and was wondering if anyone could help me to find a simpler way.

    The sum of the series 1+2+3+4+....+n= n(n+1)/2

    What is the sum of 2+6+20+....+n(n+1)=???
    I was wondering is the sum of series you are looking for actually:
    2+6+12+20+....+n(n+1)=?

    Because it's:
    1(2)+2(3)+3(4)+4(5)+...+n(n+1)
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  3. #3
    Senior Member OReilly's Avatar
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    If you are looking for sum of series:
    2 + 6 + 12 + 20 + .... + n(n + 1) =?

    then its:
    2 + 6 + 12 + 20 + .... + n(n + 1) = \frac{{n(n + 1)(n + 2)}}{3}

    You need to look for:
    2(1 + 3 + 6 + 10 + ... + \frac{{n(n + 1)}}{2}) = 2(\frac{{n(n + 1)(n + 2)}}{6}) = \frac{{n(n + 1)(n + 2)}}{3}

    Look at this post:
    http://www.mathhelpforum.com/math-he...1-2-3-4-a.html
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  4. #4
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    Quote Originally Posted by Finney View Post
    Hello I have found a pretty complicated way of finding the sum to this series, and was wondering if anyone could help me to find a simpler way.

    The sum of the series 1+2+3+4+....+n= n(n+1)/2

    What is the sum of 2+6+20+....+n(n+1)=???
    You have,
    \sum_{k=1}^n k(k+1)= \sum_{k=1}^n k^2+k= \sum_{k=1}^n k^2+\sum_{k=1}^n k
    Using some theorems,
    \frac{n(2n+1)(n+1)}{6}+\frac{n(n+1)}{2}
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  5. #5
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    Aha

    Thats very clever, the above post method I tried but it did not fork by far.
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