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Show that this sequence satisfies the recurrence relation

  1. Oct 31, 2012 #1
    1. The problem statement, all variables and given/known data
    Let d0, d1, d2,... be defined by the formula dn = 3n - 2n for all integers n ≥ 0. Show that this sequence satisfies the recurrence relation.

    dk = 5dk-1 - 6dk-2.


    2. Relevant equations



    3. The attempt at a solution

    I found that dk = 3k - 2k

    dk-1 = 3k-1 - 2k-1
    dk-2 = 3k-2 - 2k-2

    after plugging dk-1 and dk - 2 into the formula dk = 5dk-1 - 6dk-2, i am stuck and do not understand how to do the algebra if that is what i 'm supposed to be doing...any help?
     
  2. jcsd
  3. Oct 31, 2012 #2

    SammyS

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    Take
    dk-1 = 3k-1 - 2k-1

    and

    dk-2 = 3k-2 - 2k-2 '​
    Plug those into
    5dk-1 - 6dk-2 .​
    Do some algebra & see what you get.
     
  4. Oct 31, 2012 #3
    This is what i tried doing at i stated above...after you plug them in i get

    3k - 2k = 5( 3k-1 - 2k-1) - 6(3k-2 - 2k-2), I don't really understand what i can do with that algebraically...i must just be missing it...
     
  5. Oct 31, 2012 #4

    Dick

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    Use 3^(k-1)=3*3^(k-2) and 2^(k-1)=2*2^(k-2).
     
  6. Oct 31, 2012 #5
    ok so changing the equation to that gives me:

    5(3*3k-2 - 2*2k-2) - 6( 3k-2 - 2k-2)

    i can see how that gave me some like terms but multiplying through gives me:

    15 * 3k-2 - 10 * 2k-2 - 6 * 3k-2 - 6 * 2k-2

    i feel like that wasn't where you were leading me lol...i'm sorry, i'm not that great at algebra
     
  7. Oct 31, 2012 #6

    Dick

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    I guess not. But you're almost there. Collect the 3^(k-2) terms. What do you get? And you've a sign error on the last term. Could you fix it?
     
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