Results 1 to 2 of 2

Math Help - deduce the sumation of series

  1. #1
    Junior Member
    Joined
    Aug 2012
    From
    The Earth
    Posts
    71
    Thanks
    1

    deduce the sumation of series

    If f(r) = r(r+1)(r+2), simplify f(r+1)-f(r). Hence find ∑_(r=1)^n▒〖3(r^2+3r+2)〗and deduce ∑_(r=1)^n▒r^2. I stuck at the deduction part, can anyone explain how to do it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    525
    Thanks
    146

    Re: deduce the sumation of series

    Since f(r) = r(r+1)(r+2), have f(r+1)-f(r) = (r+1)((r+1)+1)((r+1)+2) - r(r+1)(r+2) = (r+1)(r+2)(r+3) - r(r+1)(r+2) = (r+1)(r+2){(r+3)-(r)} = 3(r+1)(r+2).

    Thus f(r+1)-f(r) = 3(r^{2}+3r+2).

    \sum_{r=1}^{n} 3(r^{2}+3r+2) = \sum_{r=1}^{n} ( f(r+1) - f(r) ) = f(n+1) - f(1) (the last is because the sum telescopes - everything else cancels out.)

    Now get f(n+1) - f(1): f(n+1) - f(1) = (n+1)(n+2)(n+3) - (6) = n^{3} + 6n^{2} + 11n + 6 - 6

    Thus f(n+1) - f(1) = n(n^{2} + 6n + 11)

    You're presumed to already know that \sum_{r=1}^{n} 1 = n, and \sum_{r=1}^{n} r = n(n+1)/2.

    Now \sum_{r=1}^{n} 3(r^2+3r+2) = 3 ( \sum_{r=1}^{n} r^{2} ) + 9 ( \sum_{r=1}^{n} r ) + 6 ( \sum_{r=1}^{n} 1 )

    So pluggin in gives: ( f(n+1) - f(1) ) = 3 ( \sum_{r=1}^{n} r^{2} ) + 9 ( n(n+1)/2 ) + 6 ( n )

    Solving for our desired series, \sum_{r=1}^{n} r^{2}, gives:

    \sum_{r=1}^{n} r^{2} = (1/3)( f(n+1) - f(1) -  9n(n+1)/2 - 6n )
     = (1/3)( n(n^{2} + 6n + 11) - 9n(n+1)/2 - 6n )
     = (n/3)( n^{2} + 6n + 11 - 9(n+1)/2 - 6 )
     = (n/3)( n^{2} + (3/2)n + 1/2 )
     = (n/6)( 2n^{2} + 3n + 1 )
     = (n/6)( ( n + 1 ) ( 2n + 1 ) )
     = ( n )( n + 1 )( 2n + 1 )/6

    Thus \sum_{r=1}^{n} r^{2} = ( n )( n + 1 )( 2n + 1 ) / 6
    Last edited by johnsomeone; September 10th 2012 at 06:03 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How to deduce ds/dx+ds/dy=0 from s_x(x)=0
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: November 27th 2011, 06:52 AM
  2. Deduce
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 7th 2011, 10:19 AM
  3. Deduce Q
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 26th 2009, 01:24 PM
  4. deduce
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: August 24th 2009, 09:03 PM
  5. sumation of cos x
    Posted in the Calculus Forum
    Replies: 6
    Last Post: May 2nd 2009, 11:14 PM

Search Tags


/mathhelpforum @mathhelpforum