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Homework Help: Induction Proof: Am I on the right track?

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Let a(1)=a(2)=5 and a(n+1)=a(n)+6a(n-1), n≥2
    Use induction to prove that a(n)=(3^n)-(-2)^n for n≥1

    2. Relevant equations

    Not applicable

    3. The attempt at a solution

    I have check that a(3) = 5+6·5 = 35 = 3^3-(-2)^3 so it holds for n = 3.
    The cases n = 1 and n = 2 are similar and also hold

    So I assumed that it holds for n and considered

    a(n+1) = a(n)+6a(n-1)
    = (3^n-(-2)^n)+6(3^(n-1)-(-2)^(n-1))

    The second term [6(3^(n-1)-(-2)^(n-1))] equals [2·3^n+3·(-2)^n].


    a(n+1) = (3^n-(-2)^n)+(2·3^n+3·(-2)^n)
    = 3^(n+1)-(-2)^(n+1)

    Is this a valid induction proof? Am I on the right lines here?

  2. jcsd
  3. Mar 8, 2012 #2


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    welcome to pf!

    hi elizaburlap! welcome to pf! :smile:

    (try using the X2 and X2 buttons just above the Reply box :wink:)
    yes, that's all (difficult to read :wink:, but) fine! :smile:
  4. Mar 8, 2012 #3
    Thank you! This was my first attempt at an induction proof, so I wasn't too sure.

    Oh! I see the x2 now, thanks :)
  5. Mar 10, 2012 #4
    Math1115. (:p)
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