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Another on recurrence sequences

  1. Nov 6, 2007 #1
    How would you go on proving the following conjecture?
    S_0 = 0, \quad S_1 = 1, \quad S_n = a S_{n-1} + b S_{n-2}
    Prove that
    { S_n }^2 - S_{n-1} S_{n+1} = (-b)^{n-1} \quad (n = 1, 2, 3, ...)
  2. jcsd
  3. Nov 7, 2007 #2


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    Induction seems to work. I put the first couple of steps in white (so highlight if you can't see):

    S(n+1)^2 - S(n)S(n+2)
    = S(n+1)^2 - S(n)(aS(n+1) + bS(n))
    = S(n+1)^2 - aS(n)S(n+1) - bS(n)^2

    Now think about you can possibly get a (-b)^(n-1) in there.
  4. Nov 7, 2007 #3
    Thanks! Nevermind, I was just doing something like that. Guess I was dense this morning.

    Just start from the equation on n-1 and add and substract a . S_n . S_{n-1} to the left-hand side.

    Edit: Oh, sorry, first multiply the whole equation by -b, then add and substract the term above.
    Last edited: Nov 7, 2007
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