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Can someone help me with this sequence?2 0 2/27 0

  1. Jul 30, 2010 #1
    Can someone help me with this sequence?

    2 0 2/27 0 2/125

    If I only look at the odd numbers it's: 2/n^3

    But I don't know how they get that zero..
     
  2. jcsd
  3. Jul 30, 2010 #2

    CRGreathouse

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    Re: sequence

    Could be just every other term is 0.

    A convenient way to write that would be 1 - (-1)^n in place of the 2.
     
  4. Jul 30, 2010 #3
    Re: sequence

    Thank you very much!
     
  5. Jul 30, 2010 #4
    Re: sequence

    Can you help me with this sequence also?

    1 4 1 16 1 36

    If I only look at the odd numbers it's: n^2

    But I don't know how they get to '1'..
     
  6. Jul 30, 2010 #5

    CRGreathouse

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    Re: sequence

    Again, just define it that way: f(2n + 1) = 1, f(2n) = 4n^2. You can use the same trick with (-1)^n if you want -- make the exponent 0 for odds and 1 for evens.
     
  7. Jul 30, 2010 #6
    Re: sequence

    How do you define f(n)?
     
  8. Jul 31, 2010 #7

    HallsofIvy

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    Re: sequence

    He just did. If n is odd, n= 2m+ 1 for some integer m so f(n)= 1, if n is even, n= 2m for some integer m so f(n)= f(2m)= 4(m)^2. Of course, if n= 2m then m= n/2 so you could also write this as 4(n/2)^2= 4n^2/4= n^2.
     
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