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Thread: Sequence Problem

  1. #1
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    Sequence Problem

    The Sequence $\displaystyle u_n (n = 1,2,3...) $ satisfies the recurrence relation

    $\displaystyle u_{n+2} = \frac{u_{n+1}}{u_{n}} (ku_{n} -u_{n+1})$
    where k is constant.

    If $\displaystyle u_1 = a$ and $\displaystyle u_2 = b$ where a and b are non-zero and $\displaystyle b \neq ka$

    Prove by induction that

    $\displaystyle u_{2n} = \left( \frac{b}{a} \right ) u_{2n-1}$

    $\displaystyle u_{2n+1} = cu_{2n}$

    for $\displaystyle n \geq 1$where c is a constant to be found in terms of k, a and b hence find and express $\displaystyle u_{2n}$ and $\displaystyle u_{2n-1}$ in terms of a,b,c and n

    find conditions on a, b and k in the three cases.

    i) the sequence $\displaystyle u_n$ is geometric
    ii) the sequence $\displaystyle u_n$ has period 2
    iii) the sequence $\displaystyle u_n$ has period 4

    [OCR STEP(III) 2005, Question 4 ]

    I am having some trouble on the induction part of the question, I could do with a tiny bit of guidance, I do not want a full solution just a bit of help if anyone wants to give me a hand.
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  2. #2
    Senior Member Peritus's Avatar
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    first let's start by checking the base case n = 1:

    $\displaystyle \begin{gathered}
    \left. {u_{2n} } \right|_{n = 1} = \left. {\left( {\frac{b}
    {a}} \right)u_{2n - 1} } \right|_{n = 1} \hfill \\
    \Leftrightarrow u_2 = \left( {\frac{b}
    {a}} \right)u_1 = b\quad o.k. \hfill \\
    \end{gathered} $

    now let us assume that the relation holds for some n:


    $\displaystyle
    1) u_{2n} = \left( {\frac{b}
    {a}} \right)u_{2n - 1}
    $

    now if we prove that the relation holds for n+1, then by the induction hypothesis the relation is true for any n,
    thus we have to prove that:

    $\displaystyle
    u_{2n + 2} = \left( {\frac{b}
    {a}} \right)u_{2n + 1}
    $

    our next step is to use the given recurrence relation, since it is true for every n it must be true in particular for 2n-1:

    $\displaystyle
    2)\quad u_{2n + 1} = ku_{2n} - \frac{{u_{2n} ^2 }}
    {{u_{2n - 1} }}
    $

    for the same reason it must be true for 2n, so we get:

    $\displaystyle
    3)\quad u_{2n + 2} = ku_{2n + 1} - \frac{{u_{2n + 1} ^2 }}
    {{u_{2n} }}
    $

    substitute 1) into 2) and then 2) into 3), and you;ll complete your proof...
    Last edited by Peritus; Feb 19th 2008 at 11:43 AM.
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