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Math Help - Sequence Problem

  1. #1
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    Sequence Problem

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

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

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

    Prove by induction that

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

    u_{2n+1} = cu_{2n}

    for n \geq 1where c is a constant to be found in terms of k, a and b hence find and express u_{2n} and 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 u_n is geometric
    ii) the sequence u_n has period 2
    iii) the sequence 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:

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

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


    <br />
1)   u_{2n}  = \left( {\frac{b}<br />
{a}} \right)u_{2n - 1} <br />

    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:

    <br />
u_{2n + 2}  = \left( {\frac{b}<br />
{a}} \right)u_{2n + 1} <br />

    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:

    <br />
2)\quad u_{2n + 1}  = ku_{2n}  - \frac{{u_{2n} ^2 }}<br />
{{u_{2n - 1} }}<br />

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

    <br />
3)\quad u_{2n + 2}  = ku_{2n + 1}  - \frac{{u_{2n + 1} ^2 }}<br />
{{u_{2n} }}<br />

    substitute 1) into 2) and then 2) into 3), and you;ll complete your proof...
    Last edited by Peritus; February 19th 2008 at 12:43 PM.
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