Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook