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Thread: Sum proof...need a little help

  1. #1
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    Question Sum proof...need a little help

    $\displaystyle \sum_{i=1}^{n-1} i(i+1) = (n(n-1)(n+1))/3.
    $
    For all integers $\displaystyle n \geq 2
    $

    Base case: lets n = 2

    $\displaystyle \sum_{i=1}^{n-1} i(i+1) = (2(2-1)(2+1))/3.
    $


    = 2(1)(3)
    =2

    Inductive step:

    $\displaystyle \sum_{i=1}^{(n+1)-1} i(i+1) = $$\displaystyle \sum_{i=1}^{n} i(i+1) + [(n+1)((n+1)+1)]
    $

    $\displaystyle (n(n-1)(n+1))/3 + [(n+1)(n+2)]
    $

    $\displaystyle (n(n-1)(n+1)+3(n+1)(n+2))/3
    $

    $\displaystyle ((n+1)(n^2+2n+6))/3
    $

    And this is where i am stuck cause I dont really know what to do with the
    $\displaystyle n^2+2n+6.
    $
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  2. #2
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    Take note of the special sums:
    $\displaystyle \sum\limits_{k = 1}^n k = \frac{{n(n + 1)}}{2}\;\& \,\sum\limits_{k = 1}^n {k^2 } = \frac{{n(n + 1)(2n + 1)}}{6}$

    Then write $\displaystyle \sum\limits_{k = 1}^n {k(k + 1)} = \sum\limits_{k = 1}^n {k^2 } +\sum\limits_{k = 1}^n k $.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Take note of the special sums:
    $\displaystyle \sum\limits_{k = 1}^n k = \frac{{n(n + 1)}}{2}\;\& \,\sum\limits_{k = 1}^n {k^2 } = \frac{{n(n + 1)(2n + 1)}}{6}$

    Then write $\displaystyle \sum\limits_{k = 1}^n {k(k + 1)} = \sum\limits_{k = 1}^n {k^2 } +\sum\limits_{k = 1}^n k $.

    I just substutited $\displaystyle \sum\limits_{k = 1}^n {k(k + 1)}$
    by the given statement. I dunno how those special cases would come into play for this..
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  4. #4
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    Quote Originally Posted by Kitizhi View Post
    I dunno how those special cases would come into play for this..
    How much do you know about sums?
    $\displaystyle \begin{gathered}
    \sum\limits_k {\left( {a_k + b_k } \right)} = \sum\limits_k {\left( {a_k } \right)} + \sum\limits_k {\left( {b_k } \right)} \hfill \\
    \sum\limits_k {k\left( {k + 1} \right)} = \sum\limits_k {\left( {k^2 + k} \right)} = \sum\limits_k {\left( {k^2 } \right)} + \sum\limits_k {\left( k \right)} \hfill \\
    \end{gathered} $

    Using induction is a waste of resources.
    If you are asked to study SUMS then you ought to know about SPECIAL SUMS.
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  5. #5
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    not that much just some of the basic rules, no special cases.

    I have this answer to this question and we mainly focus on the use of induction but i dont fully understand it which is why I am seeking help.
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  6. #6
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    Quote Originally Posted by Kitizhi View Post
    not that much just some of the basic rules, no special cases.
    I have this answer to this question and we mainly focus on the use of induction but i dont fully understand it which is why I am seeking help.
    Well good luck with that.
    I do think it is a pity that you are being asked to reinvent a wheel.
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