Results 1 to 5 of 5

Math Help - Finding the general term from summation

  1. #1
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Finding the general term from summation

    Given that \sum_{r = 1}^{n}U_r=3n^2+4n, find U_n and \sum_{r = n+1}^{2n}U_r
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by Punch View Post
    Given that \sum_{r = 1}^{n}U_r=3n^2+4n, find U_n and \sum_{r = n+1}^{2n}U_r
    U_n=\left(\sum_{r = 1}^{n}U_r\right)-\left(\sum_{r = 1}^{n-1}U_r\right).

    Now you show us some effort.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Quote Originally Posted by Plato View Post
    U_n=\left(\sum_{r = 1}^{n}U_r\right)-\left(\sum_{r = 1}^{n-1}U_r\right).

    Now you show us some effort.
    U_n=\left(\sum_{r = 1}^{n}U_r\right)-\left(\sum_{r = 1}^{n-1}U_r\right)=(3n^2+4n)-(3(n-1)^2+4(n-1))
    =3n^2+4n-(3n^2-6n+3+4n-4)
    =6n+1

    thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Very good.
    Now what is U_{2n}-U_n~?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2011
    Posts
    169
    9n^2+4n
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 26th 2011, 02:33 AM
  2. General term
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 10th 2010, 11:21 PM
  3. general term
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 15th 2009, 03:00 AM
  4. general term
    Posted in the Algebra Forum
    Replies: 5
    Last Post: September 2nd 2009, 07:11 AM
  5. Replies: 7
    Last Post: August 31st 2007, 08:18 PM

Search Tags


/mathhelpforum @mathhelpforum