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Recurrence relations discrete math problem

  1. May 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the general solution to the following recurrence relations (defined n≥2).

    c) an=6an-1-9an-2+8n+4



    2. Relevant equations



    3. The attempt at a solution

    an=6an-1-9an-2+8n+4

    8n+4= an -6an-1+9an-2

    R2-6R+9=0

    R=3,3

    So hn=A(3)n+B(3)n

    Assume pn=Cn+Cn2 → This is where I got stuck!

    What should my assumption be??

    I thought this because RHS =8n+4

    so for the 4 I used Cn since 4 is a constant so I put an extra n term...likewise for 8n.

    but it turned out to be a mess!

    HELP
     
  2. jcsd
  3. May 16, 2012 #2

    tiny-tim

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    hi charmedbeauty! :smile:
    no, if a recurrence relation has a root R repeated k times, the k independent solutions are niRn (0≤i<k) :wink:
     
  4. May 16, 2012 #3
    hmm im a little confused what should i be?

    so i replace hn=A(3)n+B(3)n

    with hn=A(3n)+Bn(3n)??
     
  5. May 17, 2012 #4

    tiny-tim

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    yup! :smile:

    (and you can leave out those brackets … A3n + Bn3n :wink:)
     
  6. May 17, 2012 #5
    Thanks.
     
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