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Sequence 2, 5, 8, 11, 14

  1. Feb 12, 2006 #1
    This is my question: Which powers of numbers in the sequence below are always in the sequence and which are not. Prove it?

    Sequence: 2, 5, 8, 11, 14...

    So the gerenal term is 3n + 2

    (3n+2)^2 = 9n^2 + 12n +4

    Where should I go from here?
    Last edited: Feb 12, 2006
  2. jcsd
  3. Feb 12, 2006 #2

    I think that should be 9n^2 + 12n + 4
  4. Feb 12, 2006 #3


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    Conventionally, [itex]\mathbb{N}[/itex] is the index set of sequence. This means that when identifying the general term, it must be that a1 is the first term in the sequence. The way you wrote an, a0 is your first term.

    It's no big deal, it just avoids confusion.

    I suggest you go back to your iniital problem before tacking this one as the method of proof is very similar and you're just one step away from the final solution in the other problem.
  5. Feb 12, 2006 #4


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    When you expand (3n + 2)^k into a polynomial, every term except for the constant term must be divisible by 3.
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