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Homework Help: Sequence Question

  1. Mar 16, 2010 #1
    1. The problem statement, all variables and given/known data
    U( r ) = r(r+1)(r+2)(r+3), show that U (r + 1) - U ( r ) = 4(r+1)(r+2)(r+3) Hence, find the sum to n terms of the series 2x3x4 + 3x4x5 + 4x5x6 +.....



    2. Relevant equations

    Sequence Knowledge..

    3. The attempt at a solution

    I can't even TOUCH on the problem. Can you please help?
     
  2. jcsd
  3. Mar 16, 2010 #2

    tiny-tim

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    Hi l46kok! :smile:
    I assume you can do that?
    ok, define U(r + 1) - U( r ) = V(r).

    Then the question is asking you for 1/4 ∑ V(r).

    Hint: what is V(r) + V(r+1), in terms of Us ? :wink:
     
  4. Mar 16, 2010 #3
    U( r ) = r(r+1)(r+2)(r+3), show that U (r + 1) - U ( r ) = 4(r+1)(r+2)(r+3)

    U(r+1) is = (r+1)(r+2)(r+3)(r+4)

    so U(r+1) - U( r ) = (r+1)(r+2)(r+3)(r+4) - r(r+1)(r+2)(r+3)

    let (r+1)(r+2)(r+3) = z

    = z((r+4)-r)

    = 4z

    = 4(r+1)(r+2)(r+3)



    Then if we say V(r) = U(r+1) - U( r )

    V(r) + V(r + 1) = U(r+1) - U(r) + U(r+2) - U(r+1)

    = U(r+2) - U(r)

    Then we're finding 1/4 of Summation of V(r) so

    Summation of 0.25V(r) = 0.25 U(r+n) - U(r)

    ?? is this it?

    oh yeah and where did the 1/4 come from?
     
    Last edited: Mar 16, 2010
  5. Mar 16, 2010 #4

    HallsofIvy

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    Excellent!



    Since V(r)= U(r+1)- U(r), V(1)+ v(2)= U(2)- U(1)+ U(3)- U(2)= U(3)- U(1).

    V(1)+ V(2)+ V(3)= U(2)- U(1)+ U(3)- U(2)+ U(4)- U(3)= U(4)- U(1).

    V(1)+ V(2)+ V(3)+ V(4)= U(2)- U(1)+ U(3)- U(2)+ U(4)- U(3)+ U(5)- U(4)= U(5)- U(1).

    Get the point? This is a "telescoping series" [itex]\sum_{i=1}^n V(i)= U(n+1)- U(1)[/iyrc].

    The "1/4" is to get rid of the "4" in the formula for V(r). Since V(r)= 4(r+1)(r+2)(r+3), the sum you are asked to do is (1/4)(V(1)+ V(2)+ ...+ V(n)).
     
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