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Find the sum of a series.

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Hello there!

    I'm trying to find the sum of:

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

    3. The attempt at a solution

    Well, I tried with

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

    with x = 3. However, the nth term doesn't fit (3^(2n)).

    At the moment, I'm a little bit out of ideas. Any suggestion would be very appreciated.
     
  2. jcsd
  3. Feb 15, 2012 #2

    Dick

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    This formula is wrong. Why don't you look up the correct one to get started?
     
  4. Feb 15, 2012 #3
    Oh yeah, my bad.

    I meant

    1 + x + x^2 + x^3 + ... = (1 - x^n)/(1 - x)
     
  5. Feb 15, 2012 #4

    Dick

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    Still not good. There should be a last term on the left. It's not an infinite series.
     
  6. Feb 15, 2012 #5
    1 + x + x^2 + x^3 + ... + x^(n-1) = (1 - x^n)/(1 - x)

    I'm currently trying, but still nothing.
     
  7. Feb 15, 2012 #6

    Dick

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    That's it. Let's write that as 1 + x + x^2 + x^3 + ... + x^(m-1) = (1 - x^m)/(1 - x). It looks to me like you just want x=(-3) and m-1=2n.
     
  8. Feb 15, 2012 #7
    I'm trying for example:

    The nth term 3^(2n), I'm wrinting it as (3^n)^2

    Then I got: (Sigma 3^n) times (Sigma 3^n). Each one is a geometric series and the sum for each would be (1 - 3^n)/(1 - 3). Since there is a product of geometric series, I supposed that on the RHS I could put (1 - 3^n)/(1 - 3) times (1 - 3^n)/(1 - 3).

    Or [(1 - 3^n)/(1 - 3)]^2

    But apparently it doesn't work...
     
  9. Feb 15, 2012 #8

    Dick

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    No. It's a single geometric series. You hardly have to do anything except substitute into a formula. 3^(2n)=(-3)^(2n). Reread my last post.
     
  10. Feb 15, 2012 #9
    Omg. I think my mistake was that I used all the time x = 3 and not x =-3!

    Let's see.
     
  11. Feb 15, 2012 #10
    ...

    Let's see:

    m- 1 = 2n, m = 2n + 1, and x = -3,

    (1 - (-3^{2n+1})/ 4

    For n = 0, 1.

    [1 - (-3^1)]/4 = 1, [1 - (-3^3)]/4 = 28/4 = 7.

    but the sum of the first two terms is -2, and not 7.

    1 - 3 + 3^2 - 3^3 + ... + 3^(2n)

    Apparently doesn't work, (or its me?) Why it doesn't work?
     
  12. Feb 15, 2012 #11

    Dick

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    If you put n=1 then you are summing up to 3^(2*n)=3^2=9. That's the sum of the first three terms 1-3+9. Not the first two terms.
     
  13. Feb 15, 2012 #12
    Aha! Now I get it. Thank you very much for your help!
     
  14. Feb 15, 2012 #13
    Now, I'm trying to prove this by induction, but I doesn't work...

    [1 - x^{2n+1}/4] + 3^{2n+1} = [1 - x^{2n+1} + 4x^{2n+1}]/4

    [1 - x^{2n+1} (1 + 4)]/4 = [1 - 5x^{2n+1}]/4,

    susbtituting x =-3, [1 - 5(-3)^{2n+1}]/4,

    but, I should get [1 - (-3)^{2n+3}/4]

    Any suggestion?
     
  15. Feb 15, 2012 #14

    Dick

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    The n case is the sum up to (-3)^(2n). To get to the n+1 case, which sums up to (-3)^(2(n+1))=(-3)^(2n+2) you need to add two terms. (-3)^(2n+1) and (-3)^(2n+2).
     
    Last edited: Feb 15, 2012
  16. Feb 15, 2012 #15
    Yeah, indeed, it works.

    I must admit I don't fully understand why for the n+1 term, I must add two terms. Why is that?
     
  17. Feb 15, 2012 #16
    So, from the n+1 term: (-3)^(2(n+1))=(-3)^(2n+2)

    the distance from n, is 2n+2, and therefore I must include the number which is in between, that is, 2n+1?
     
  18. Feb 15, 2012 #17

    Dick

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    Yes. The case n and the case n+1 differ by the addition of two terms.
     
  19. Feb 15, 2012 #18
    Ok. Thanks again!
     
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