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Proof by induction sequence

  1. Feb 10, 2014 #1
    Define the sequence of integers a1, a2, a3,.... as follows:
    a1 = 3
    a2 = 6
    an = 5an-1 - 6an-2 + 2 for all n ≥ 3
    Prove that an = 1 + 2n-1 + 3n-1

    ------------------------------------------------------------------------------------------------
    my attempt:
    base case:
    n=1
    1+ 20 +30
    = 1
    = a1
    therefore n=1 holds

    n=2
    1+ 21 +31
    = 6
    = a2
    therefore n=2 holds

    Assume k holds for all k > n, n≥3 then

    an = 5an-1 - 6an-2 + 2
    = 5(1+2n-2 + 3n-2) -6(1 + 2n-3+ 3n-3) +2
    = 1 + 10n-2 + 15n-2 -12n-3 -18n-3

    I have no idea what to do from here.
     
    Last edited: Feb 10, 2014
  2. jcsd
  3. Feb 10, 2014 #2

    eumyang

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    [itex]5 \cdot 2^{n-2} \ne 10^{n-2}[/itex]
    [itex]5 \cdot 3^{n-2} \ne 15^{n-2}[/itex]
    ...and so on.

    I don't know if this will help, but note that
    [itex]2^{n-2} = 2\cdot 2^{n-3}[/itex]
    and
    [itex]3^{n-2} = 3\cdot 3^{n-3}[/itex]
     
  4. Feb 10, 2014 #3

    Mentallic

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    Your second line is incorrect.

    [tex]a\cdot b^n = a^1b^n \neq (ab)^n = a^nb^n[/tex]
     
  5. Feb 11, 2014 #4
    okay, but I am still stuck
     
  6. Feb 11, 2014 #5

    eumyang

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    Best I can do is give you some other examples that use the properties that you may need to continue:
    [itex]4(2+5^{n-3}) = 8 + 4 \cdot 5^{n-3}[/itex]
    [itex]-7 \cdot 4^{n-1} = -7 \cdot 4 \cdot 4^{n-2} = -28 \cdot 4^{n-2}[/itex]
    [itex]12 \cdot 10^{n-5} - 4 \cdot 10^{n-5} = 8 \cdot 10^{n-5}[/itex]

    Study the above and try your problem again.


    (Mods: hope I am not giving too much away here.)
     
  7. Feb 11, 2014 #6

    haruspex

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    Please post your corrected working, as far as it goes. (Maybe you still have a wrong equation.)
     
  8. Feb 11, 2014 #7
    I found the answer, thank you all so much!
     
    Last edited: Feb 11, 2014
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