Results 1 to 7 of 7
Like Tree1Thanks
  • 1 Post By Wilmer

Math Help - Mathmatical induction

  1. #1
    Newbie
    Joined
    Nov 2011
    Posts
    10

    Exclamation Mathmatical induction

    how do i get from

    (1/2) - (1/k(k+1)) + (2/(k+1)(k+2)

    to (1/2) - (1/(k+1)(k+2)

    Any help would be great!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,513
    Thanks
    769

    Re: Mathmatical induction

    What do you mean by "get from ... to ..."? The equality of these two expressions seems to hold only for k = 1: WolframAlpha.

    Could you post the original problem?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,094
    Thanks
    67

    Re: Mathmatical induction

    Bad bracketing...repost clearly...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,406
    Thanks
    1294

    Re: Mathmatical induction

    Quote Originally Posted by gurnster76 View Post
    how do i get from

    (1/2) - (1/k(k+1)) + (2/(k+1)(k+2)

    to (1/2) - (1/(k+1)(k+2)

    Any help would be great!
    Why is this in Geometry when it's clearly Algebra? Anyway, it doesn't...

    \displaystyle \begin{align*} \frac{1}{2} - \frac{1}{k(k + 1)} + \frac{2}{(k + 1)(k + 2)} &= \frac{1}{2} - \frac{k + 2}{k( k + 1)(k + 2)} + \frac{2k}{k(k + 1)(k + 2)} \\ &= \frac{1}{2} + \frac{2k - k - 2}{k(k + 1)(k + 2)} \\ &= \frac{1}{2} + \frac{k - 2}{k(k + 1)(k + 2)} \end{align*}

    which does NOT equal \displaystyle \begin{align*} \frac{1}{2} - \frac{1}{(k + 1)(k + 2)} \end{align*}

    I expect you might have made a mistake somewhere in your induction process, which is why it's advisable to always post the ENTIRE question and ALL of your working...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2011
    Posts
    10

    Re: Mathmatical induction

    sorry for the bad bracketing, this is the original question, im still massively confused by this.

    Use mathematical induction to prove that


    n
    Σ 2/r(r^2-1) = (1/2) - (1/n(n+1)) n= 2,3,4.......
    r=2

    i have done the basic proof and got to the point


    k+1
    Σ = (1/2) - (1/k(k+1)) + 2/(k+1)(k+1^2-1)
    r=2


    I think i need to get to (1/2) - 1/(k+1)(k+2) but im not sure how to get there, any help would be greatly appreciated!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1

    Re: Mathmatical induction

    Quote Originally Posted by gurnster76 View Post
    sorry for the bad bracketing, this is the original question, im still massively confused by this.
    Use mathematical induction to prove that
    n
    Σ 2/r(r^2-1) = (1/2) - (1/n(n+1)) n= 2,3,4.......
    r=2
    This is a good lessen: ALWAYS, ALWAYS post the entire exact question.
    By not doing so, you have a wasted much of time.

    \sum\limits_{r = 2}^n {\frac{2}{{r\left( {{r^2} - 1} \right)}}}  = \frac{1}{2} - \frac{1}{{n\left( {n + 1} \right)}}

    Now you want to go from \frac{1}{2} - \frac{1}{{n\left( {n + 1} \right)}} + \frac{2}{{\left( {n + 1} \right)\left( {{{\left( {n + 1} \right)}^2} - 1} \right)}} to \frac{1}{2} - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,094
    Thanks
    67

    Re: Mathmatical induction

    Quote Originally Posted by gurnster76 View Post
    2/r(r^2-1)
    That means 2 divided by r, then the result multiplied by (r^2-1)

    Do you mean 2 divided by the whole thing; if so, this is proper way: 2/(r(r^2-1))
    Thanks from emakarov
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mathmatical modeling
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: November 21st 2011, 02:06 AM
  2. Replies: 10
    Last Post: June 29th 2010, 12:10 PM
  3. Dishwashing mathmatical help
    Posted in the Algebra Forum
    Replies: 7
    Last Post: June 6th 2010, 07:32 AM
  4. Is there a mathmatical solution to this problem?
    Posted in the Advanced Math Topics Forum
    Replies: 6
    Last Post: September 1st 2007, 04:44 PM
  5. Another Mathmatical Problem..
    Posted in the Advanced Statistics Forum
    Replies: 9
    Last Post: August 8th 2006, 03:59 AM

Search Tags


/mathhelpforum @mathhelpforum