Results 1 to 6 of 6

Math Help - Sum proof...need a little help

  1. #1
    Junior Member
    Joined
    Feb 2009
    Posts
    48

    Question Sum proof...need a little help

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

    Base case: lets n = 2

    \sum_{i=1}^{n-1} i(i+1) = (2(2-1)(2+1))/3.<br />

    = 2(1)(3)
    =2

    Inductive step:

    \sum_{i=1}^{(n+1)-1} i(i+1) = \sum_{i=1}^{n} i(i+1) + [(n+1)((n+1)+1)]<br />

    (n(n-1)(n+1))/3 + [(n+1)(n+2)]<br />

    (n(n-1)(n+1)+3(n+1)(n+2))/3<br />

    ((n+1)(n^2+2n+6))/3<br />

    And this is where i am stuck cause I dont really know what to do with the
    n^2+2n+6.<br />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    Take note of the special sums:
    \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 \sum\limits_{k = 1}^n {k(k + 1)}  = \sum\limits_{k = 1}^n {k^2 }  +\sum\limits_{k = 1}^n k .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2009
    Posts
    48
    Quote Originally Posted by Plato View Post
    Take note of the special sums:
    \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 \sum\limits_{k = 1}^n {k(k + 1)}  = \sum\limits_{k = 1}^n {k^2 }  +\sum\limits_{k = 1}^n k .

    I just substutited \sum\limits_{k = 1}^n {k(k + 1)}
    by the given statement. I dunno how those special cases would come into play for this..
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    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?
    \begin{gathered}<br />
  \sum\limits_k {\left( {a_k  + b_k } \right)}  = \sum\limits_k {\left( {a_k } \right)}  + \sum\limits_k {\left( {b_k } \right)}  \hfill \\<br />
  \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 \\ <br />
\end{gathered}

    Using induction is a waste of resources.
    If you are asked to study SUMS then you ought to know about SPECIAL SUMS.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2009
    Posts
    48
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 19th 2010, 10:50 AM
  2. Replies: 0
    Last Post: June 29th 2010, 08:48 AM
  3. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 27th 2010, 10:07 PM
  4. Proof with algebra, and proof by induction (problems)
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: June 8th 2008, 01:20 PM
  5. proof that the proof that .999_ = 1 is not a proof (version)
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: April 14th 2008, 04:07 PM

Search Tags


/mathhelpforum @mathhelpforum