Results 1 to 3 of 3

Math Help - summation and mathematical induction

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    53

    summation and mathematical induction

    1.Prove by induction that \sum_{r=1}^{n}(\frac{1}{r(r+1)(r+2)}) = \frac{n(n+3)}{4(n+1)(n+2)}.

    Show that \frac{n(n+3)}{4(n+1)(n+2)} is LESSER than \frac{1}{4} for all positive integer values of n.

    Deduce from these results that \sum_{r=1}^{n}(\frac{1}{(r+1)^3}) is LESSER than \frac{1}{4}.


    i can do the first part of the question. i need help for the second and third parts in getting started.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by wintersoltice View Post
    1.Prove by induction that \sum_{r=1}^{n}(\frac{1}{r(r+1)(r+2)}) = \frac{n(n+3)}{4(n+1)(n+2)}.

    Show that \frac{n(n+3)}{4(n+1)(n+2)} is LESSER than \frac{1}{4} for all positive integer values of n.

    Deduce from these results that \sum_{r=1}^{n}(\frac{1}{(r+1)^3}) is LESSER than \frac{1}{4}.


    i can do the first part of the question. i need help for the second and third parts in getting started.
    For the second part

    \frac{1}{4}\left(\frac{n(n+3)}{(n+1)(n+2)}\right)<  \frac{1}{4} if n(n+3)<(n+1)(n+2)

    n^2+3n<n^2+3n+2\ ? yes, it is
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by wintersoltice View Post
    1.Prove by induction that \sum_{r=1}^{n}(\frac{1}{r(r+1)(r+2)}) = \frac{n(n+3)}{4(n+1)(n+2)}.

    Show that \frac{n(n+3)}{4(n+1)(n+2)} is LESSER than \frac{1}{4} for all positive integer values of n.

    Deduce from these results that \sum_{r=1}^{n}(\frac{1}{(r+1)^3}) is LESSER than \frac{1}{4}.


    i can do the first part of the question. i need help for the second and third parts in getting started.
    The third part involves showing that

    \frac{1}{(r+1)^3}<\frac{1}{r(r+1)(r+2)}

    \frac{1}{(r+1)^2}<\frac{1}{r(r+2)}\ ?

    \frac{1}{r^2+2r+1}<\frac{1}{r^2+2r}\ ?

    r^2+2r+1>r^2+2r\ ?


    The induction proof appears more challenging than parts 2 and 3.
    Here's how I would give the Induction Proof....

    P(k)

    \sum_{r=1}^{k}\frac{1}{r(r+1)(r+2)}=\frac{k(k+3)}{  4(k+1)(k+2)}\ ?

    P(k+1)

    \sum_{r=1}^{k+1}\frac{1}{r(r+1)(r+2)}=\frac{(k+1)(  k+4)}{4(k+2)(k+3)}\ ?

    Try to prove that P(k) being true causes P(k+1) to also be true.

    Proof

    \sum_{r=1}^{k+1}\frac{1}{r(r+1)(r+2)}=\sum_{r=1}^{  k}\frac{1}{r(r+1)(r+2)}+\frac{1}{(k+1)(k+2)(k+3)}

    =\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+  3)}

    if P(k) is true.

    This gives

    \frac{k(k+3)^2}{4(k+1)(k+2)(k+3)}+\frac{4}{k+1}\le  ft(\frac{1}{4(k+2)(k+3)}\right)

    \frac{1}{4(k+2)(k+3)}\left(\frac{k(k+3)^2}{k+1}+\f  rac{4}{k+1}\right)

    =\frac{1}{4(k+2)(k+3)}\left(\frac{k\left(k^2+6k+9\  right)+4}{k+1}\right)

    =\frac{1}{4(k+2)(k+3)}\left(\frac{k^3+6k^2+9k+4}{k  +1}\right)

    =\frac{1}{4(k+2)(k+3)}\frac{(k+1)\left(k^2+5k+4\ri  ght)}{k+1}

    =\frac{(k+4)(k+1)}{4(k+2)(k+3)}

    Then test your initial value to complete the proof.
    Last edited by Archie Meade; July 15th 2010 at 08:58 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. summation Induction
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 23rd 2011, 02:21 AM
  2. summation Induction
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: December 23rd 2011, 01:55 AM
  3. Replies: 10
    Last Post: June 29th 2010, 12:10 PM
  4. Summation Induction help
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: March 18th 2010, 05:20 PM
  5. Induction - Summation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 24th 2009, 04:12 AM

Search Tags


/mathhelpforum @mathhelpforum