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Solving a Linear Homogeneous Recurrence Relation

  1. Dec 12, 2011 #1
    1. Solve the following recurrence relation.
    an - 5an-1 + 6an-2 = 0, n ≥ 2, a0 = 1, a1 = 3







    3. My attempt
    I changed it to 0 = tn - 5tn-1 + 6tn-2
    Don't know where to go from there.
     
    Last edited by a moderator: Dec 12, 2011
  2. jcsd
  3. Dec 12, 2011 #2

    Office_Shredder

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    Do you see something you can factor out of that polynomial?
     
    Last edited by a moderator: Dec 12, 2011
  4. Dec 12, 2011 #3
    I'm not sure how to factor anything with (n-1) or (n-2) as an exponent.

    The book turned a similar equation into t2-t-6.
    Which made no sense to me.
     
    Last edited by a moderator: Dec 12, 2011
  5. Dec 12, 2011 #4

    Dick

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    First factor out t^(n-2). It divides all three terms, right?
     
    Last edited by a moderator: Dec 12, 2011
  6. Dec 12, 2011 #5
    Ok so I now have it down to
    t=2,3.
    Now I assume i need to put that into an equation for the original problem.
    Which I have no clue how to do.

    But where does the n≥2 stand?
    Does that affect my final outcome?
     
    Last edited by a moderator: Dec 12, 2011
  7. Dec 12, 2011 #6

    Dick

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    Are you sure you've seen examples of this before? They should have given you one before they gave you this problem. The t=2,3 tells you the solution is a_n=A*2^n+B*3^n. You know a0 and a1. Use those to find A and B.
     
  8. Dec 12, 2011 #7
    I've only seen one example.
    sn = sn-1 + 6sn-2, s0 = 4 s1 = 7

    However this equation put a 0 in place of the an and added the n≥2.
    Which maybe I shouldn't have paid attention to.

    But my final answer is: A=1 B=0. So an = 3n. Is this correct?
     
  9. Dec 12, 2011 #8

    Dick

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    It sure is.
     
  10. Dec 12, 2011 #9
    Thank you very much for your help.
     
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