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Sequence mathematics help

  1. Feb 16, 2006 #1
    Could anyone just run through my answer to this question and spote any mistakes, to let me know if I have done this correctly please? Thanks!


    Which powers of numbers in the sequence are always in the sequence and which are not. Prove your findings.

    sequence 2, 5, 8, 11, 14...


    a_n = 3n+2

    So a member in the sequence taken to a power of a number in the sequence will be of the form (3n+2)^x where x is any member of the sequence.


    (3n+2)^x = (3n)^x + (3n)^x-1 *2^1 + (3n)^x-2 * 2^2 +...+(3n)^1 * 2^x-1 + 2^x

    Noticing from this that the last term is always 2^x.

    If we set (3x+2)^x = 3q+2

    Solving for q:

    q= (2^x - 2)/3

    As q has to be an integer, x has to take values for which (2^x - 2)/3 remains an integer are the odd numbers (1, 3, 5, 7, ...)

    Therefore, the powers of the numbers which will always be in the sequence are the odd numbers. All the even powers do not lie in the sequence.
  2. jcsd
  3. Feb 16, 2006 #2


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    Homework Helper

    You've expanded the terms out incorrectly. Where are all the binomial coefficients?
    [tex](x + y) ^ n = \sum_{k = 0} ^ n \left( \begin{array}{l} n \\ k \end{array} \right) x ^ k y ^ {n - k}[/tex]
    Nope, this is wrong.
    [tex]q = \frac{(3x + 2) ^ x - 2}{3}[/tex]
    Expand (3x + 2)x out like above, we notice that all but the last term is divisible by 3, we have:
    [tex]q = \frac{(3x + 2) ^ x - 2}{3} = A + \frac{2 ^ n - 2}{3}[/tex], where A is some positive integer.
    Why? Can you prove that: 22n + 1 - 2, where n is an integer is divisible by 3?
    Hint, another way is that:
    [tex]a_n \equiv -1 \mbox{ mod } 3[/tex]
    [tex]\Rightarrow (a_n) ^ {2k} \equiv ? \mbox{ mod } 3, \ k \in \mathbb{N}[/tex]
    [tex]\Rightarrow (a_n) ^ {2k + 1} \equiv ? \mbox{ mod } 3, \ k \in \mathbb{N}[/tex]
    Can you go from here? :)
    Last edited: Feb 16, 2006
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