C2 Geometric Series

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  1. Julii92's Avatar
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    C2 Geometric Series
    The sum of the geometric series 1+r+r2+... is k times the sum of the series 1-r+r2..., where k>0. Express r in terms of k.

    By equating the sums of both series, I've got (1-rn)/(1-r) = k(1-(-r)n)/(1+r), but I'm struggling from here. Help appreciated.
  2. BabyMaths's Avatar
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    Re: C2 Geometric Series
    I think you should be looking at the sum to infinity.
  3. Julii92's Avatar
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    Re: C2 Geometric Series
    (Original post by BabyMaths)
    I think you should be looking at the sum to infinity.
    The question doesn't say anything about the value of r - would you think I could assume that -1<r<1?
    Thanks
  4. TrueGrit's Avatar
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    Re: C2 Geometric Series
    If |r| > 1 then the sum of the series is infinity, so presumable it means not.

    Also are you sure about the second series: 1-r+r2 the 'minus' ?
    Last edited by TrueGrit; 22-06-2012 at 21:00.
  5. Imposition's Avatar
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    Re: C2 Geometric Series
    (Original post by TrueGrit)
    If |r| > 1 then the sum of the series is infinity, so presumable it means not.

    Also are you sure about the second series: 1-r+r2 the 'minus' ?
    Might mean you multiply by -r
  6. TrueGrit's Avatar
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    Re: C2 Geometric Series
    (Original post by Imposition)
    Might mean you multiply by -r
    I don't know I didn't write the question...
  7. aznkid66's Avatar
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    Re: C2 Geometric Series
    (Original post by TrueGrit)
    I don't know I didn't write the question...
    More like the question most likely means a geometric series with a ratio of -r. If the minus were a plus, there wouldn't be much of a question.
  8. Imposition's Avatar
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    Re: C2 Geometric Series
    (Original post by TrueGrit)
    I don't know I didn't write the question...
    I meant that the common ratio is -r
    I think the OP's missing a part of the question out, sum to infinity maybe.
  9. Julii92's Avatar
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    Re: C2 Geometric Series
    (Original post by Imposition)
    I meant that the common ratio is -r
    I think the OP's missing a part of the question out, sum to infinity maybe.
    Just re-checked the question, it's actually written "the sum of the infinite geometric series 1+r+..." but I'm not sure what it means by infinte series.
  10. BabyMaths's Avatar
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    Re: C2 Geometric Series
    So, it took you 13 hours to get around to reading the question properly.

    :spank:

    If |r|<1 then as you add terms the sum approaches some particular value. For example S = 1+1/2+1/4+1/8+1/16......

    The sums are

    1
    1+1/2=3/2
    1+1/2+1/4=7/4
    1+1/2+1/4+1/8=15/8

    and it's clearly approaching 2 and you can get as close to 2 as you like.
    Last edited by BabyMaths; 23-06-2012 at 08:43.
  11. ztibor's Avatar
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    Re: C2 Geometric Series
    (Original post by Julii92)
    The sum of the geometric series 1+r+r2+... is k times the sum of the series 1-r+r2..., where k>0. Express r in terms of k.

    By equating the sums of both series, I've got (1-rn)/(1-r) = k(1-(-r)n)/(1+r), but I'm struggling from here. Help appreciated.
    For infinite geometric series
    \displaystyle 1+r+r^2+... =\sum_{n=0}^{\infty} r^n=\frac{1}{1-r} for |r|<1
    \displaystyle 1-r+r^2-r^3+....=\sum_{n=0}^{\infty} (-1)^n \cdot r^n=\sum_{n=0}^{\infty} (-r)^n=\frac{1}{1-(-r)}
    So your equation
    \displaystyle \frac{1}{1-r}=k\cdot \frac{1}{1+r}
    Solve for r
    then from |r|<1 determine which integer can be k (maybe it's only my question)
    Last edited by ztibor; 24-06-2012 at 09:36.
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