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Sequence and series

  1. Feb 14, 2006 #1
    Right here is my sequence 2, 5, 8, 11, 14, ....

    I have been asked to prove that the cube of any number in the sequence is in the sequence.

    my answer:

    General term: a_n=3n+2

    We need to cube a_n and see if it matches a number in the series i.e. (a_n)^3 = 3q+2 for some integer q.

    (a_n)^3
    =27n^3 + 54n^2 + 36n + 8
    =3(9n^3 + 18n^2 + 12n + 2) +2
    =3k+2

    If this is a member of the series, then 3q+2 = 3k+2 for some integer q.

    Solving for q:

    q = k which is always in the sequence.

    So the cube of any number is in this sequence.

    But now I'm asked to show which cube numbers (therefore not in the sequence, I think :confused: ) are not in the sequence and to prove it?

    A little confused how to do this one could anyone help please :-)
     
    Last edited: Feb 14, 2006
  2. jcsd
  3. Feb 14, 2006 #2

    Fermat

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    Check your expansion of (3n+2)^3

    (3n+2)^3 = (3n)^3 + 3.(3n)^2.2 + 3.(3n).2^2 + 2^3
    (3n+2)^3 = 27n^3 + 54N^2 + 36n + 8
     
  4. Feb 14, 2006 #3
    oops thanks ever so much!
     
    Last edited: Feb 14, 2006
  5. Feb 14, 2006 #4
    Can anyone see through this one? :bugeye:
     
    Last edited: Feb 14, 2006
  6. Feb 14, 2006 #5

    Fermat

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    Your sequence is a_n = 3n + 2

    ergo 3n and 3n+1 are not in the sequence.

    Does that help?
     
    Last edited: Feb 14, 2006
  7. Feb 14, 2006 #6
    well spotted! merci!
     
  8. Feb 14, 2006 #7

    Fermat

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    c'est rien!
     
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