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Math Help - recurrence problem...

  1. #1
    Newbie
    Joined
    Jul 2009
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    Red face recurrence problem...

    do u have any idea on this?

    question :


    let n >=2, 0 < X1< X2 <....< Xn-1 < Xn,
    for Xk , k=1,2,....,n to be integers

    Define
    Sn=(1 + 1/X1)(1 + 1/X2)....(1 + 1/Xn) - 1,

    then
    i) Show the recurrence Sn-1 = Xn/(Xn+1) Sn - 1

    ii)If
    Sn and Sn-1 are positive integers then Sk is also a positive integer for every k=1,2,3,.....,n


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  2. #2
    MHF Contributor red_dog's Avatar
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    Medgidia, Romania
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    Thanks
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    i) I think the recurrence is not true.

    S_n+1=\left(1+\frac{1}{x_1}\right)\left(1+\frac{1}  {x_2}\right)\ldots\left(1+\frac{1}{x_n}\right)

    S_{n-1}+1=\left(1+\frac{1}{x_1}\right)\left(1+\frac{1}{  x_2}\right)\ldots\left(1+\frac{1}{x_{n-1}}\right)

    Then S_n+1=(S_{n-1}+1)\left(1+\frac{1}{x_n}\right)=\frac{x_n+1}{x_n  }(S_{n-1}+1)

    S_{n-1}=\frac{x_n}{x_n+1}(S_n+1)-1\Rightarrow S_{n-1}=\frac{x_nS_n-1}{x_n+1}

    ii) S_k=\left(1+\frac{1}{x_1}\right)\left(1+\frac{1}{x  _2}\right)\ldots\left(1+\frac{1}{x_k}\right)-1

    Every factor of the product is greater then 1, so the entire factor is greater than 1. Therefore S_k>0, \ \forall k\geq 1
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  3. #3
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    Smile tq very much red_dog

    u're absolutely right
    sorry that i did'nt type the question correctly.

    anyway,
    thanx a lot!!
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