Results 1 to 4 of 4

Math Help - [SOLVED] Method of differences problems

  1. #1
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261

    [SOLVED] Method of differences problems

    If f(r)=\frac{1}{r^2} , simpify f(r)-f(r+1) . Hence , find the sum of the first n terms of the series

     <br />
\frac{3}{1^2\times2^2}+\frac{5}{2^2\times3^2}+\fra  c{7}{3^2\times4^2}+...<br />

    My working :

    After simplifying , i got \frac{2r+1}{r^2(r+1)^2} .

    Afrer doing the partial fraction , i got

    \sum^{n}_{r=1}\frac{2r+1}{r^2(r+1)^2}=\sum^{n}_{r=  1}[\frac{1}{r^2}-\frac{1}{r+1}\frac{1}{(r+1)^2}]

    I need help on finding the sums , i am not sure how to continue from where i stopped and i am not sure how to use the method of differences here . THanks for any help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    41
    Quote Originally Posted by mathaddict View Post
    If f(r)=\frac{1}{r^2} , simpify f(r)-f(r+1) . Hence , find the sum of the first n terms of the series

     <br />
\frac{3}{1^2\times2^2}+\frac{5}{2^2\times3^2}+\fra  c{7}{3^2\times4^2}+...<br />

    My working :

    After simplifying , i got \frac{2r+1}{r^2(r+1)^2} .

    Afrer doing the partial fraction , i got

    \sum^{n}_{r=1}\frac{2r+1}{r^2(r+1)^2}=\sum^{n}_{r=  1}[\frac{1}{r^2}-\frac{1}{r+1}\frac{1}{(r+1)^2}]

    I need help on finding the sums , i am not sure how to continue from where i stopped and i am not sure how to use the method of differences here . THanks for any help
    I think you have an extra term here. I think it should be

    \sum^{n}_{r=1}\frac{2r+1}{r^2(r+1)^2}=\sum^{n}_{r=  1}\frac{1}{r^2}-\frac{1}{(r+1)^2}

    and in that case

    \sum^{n}_{r=1}\frac{1}{r^2}-\frac{1}{(r+1)^2} = 1- \frac{1}{(n+1)^2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Thanks Danny , but can you elaborate more on your final step be4 getting the answer ? Did you apply the method of differences ? How did you do that ?

    Thanks again,
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    41
    Quote Originally Posted by mathaddict View Post
    Thanks Danny , but can you elaborate more on your final step be4 getting the answer ? Did you apply the method of differences ? How did you do that ?

    Thanks again,
    Sure. Writing out the terms we find
    <br />
\sum^{n}_{r=1}\frac{1}{r^2}-\frac{1}{(r+1)^2} = \left( \frac{1}{1^2} - \frac{1}{2^2} \right) + \left( \frac{1}{2^2} - \frac{1}{3^2} \right) + \left( \frac{1}{3^2} - \frac{1}{4^2} \right) + \cdots + \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)
    notice all of the terms cancel except the first of the first term and the last of the last term leaving
    <br />
\frac{1}{1^2} - \frac{1}{(n+1)^2}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Method of Differences
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 20th 2011, 10:16 PM
  2. [SOLVED] Sum of series by method of differences
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 10th 2010, 02:08 AM
  3. method of differences
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 3rd 2009, 07:27 AM
  4. Summation: Method of differences.
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: May 20th 2008, 04:06 AM
  5. Method of Differences
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 11th 2008, 11:28 PM

Search Tags


/mathhelpforum @mathhelpforum