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Really silly Geometric Progression question

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data
    I should know this (it's been a few years) but can't seem to get the answer for: Ʃ (from i = 0 to n) 0.5-i the answer is apparently 2n+1-1 / 2 - 1

    how do I go about getting this?

    2. Relevant equations

    I tried Ʃn-1k=0ark = a* 1-rn / 1 - r

    with no luck

    3. The attempt at a solution

    = a* 1-rn+1 / 1 - r (which is wrong)


    Thanks very much!
     
    Last edited: Nov 13, 2012
  2. jcsd
  3. Nov 13, 2012 #2

    SammyS

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    First of all, you need to use parentheses to say what (I hope) you intend to say.

    Start by letting [itex]\displaystyle \ S=\sum_{i=0}^{n}\left(\frac{1}{2}\right)^{-i}\ .[/itex]

    Of course that means that [itex]\displaystyle \ S=\sum_{i=0}^{n}\left(2\right)^{i}\ .[/itex]

    Then take 2S - S . What do you get?
     
  4. Nov 13, 2012 #3
    Thanks for the reply SammyS, sorry about the lack of brackets, you were almost correct except I didn't have an n in the exponent of the denominator and I made up for the n-1 in the summation limit by adding n+1 to the n in the numerator (for the attempted solution)

    Aaah, YES, that almost seems quite lateral in thought.


    H'mm, not sure I'm following...'S'...?
    Anyway, that equation I tried before now works if I'm not mistaken.

    BUT this begs the question, what if rather than being a half (0.5) and being able to remove the numerator making the exponent positive, what if it was like pi ? what would we do then??

    Thanks!
     
    Last edited: Nov 13, 2012
  5. Nov 13, 2012 #4

    SammyS

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    I thought that 2S - S might trip you up.

    To elaborate...

    Write S as: [itex]\displaystyle \ S=1+\sum_{i=1}^{n}\left(2\right)^{i}\ .[/itex]

    Write 2S as: [itex]\displaystyle \ 2S=\sum_{i=1}^{n}\left(2\right)^{i}\ \, +\ 2^{n+1}.[/itex]

    Now, what is 2S - S ?
     
  6. Nov 13, 2012 #5
    would it just be 2S= 2n+1 - 1? Although I don't see how 2S was equal to the expression you said in the first place...?

    Cheers
     
  7. Nov 13, 2012 #6

    Mark44

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    ????
    lateral in thought?
     
  8. Nov 13, 2012 #7
    well I didn't see it.
    Feel free to input anything else...like my question about what if it was 'pi'^-i.
     
  9. Nov 13, 2012 #8

    Mark44

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    I don't see how that would make any difference.

    $$ \sum_{i = 1}^n \pi^{-1} = \sum_{i = 1}^n \frac{1}{\pi^i} = \sum_{i = 1}^n \left(\frac{1}{\pi}\right)^i$$

    It's still a finite geometric series.
     
  10. Nov 13, 2012 #9

    Ray Vickson

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    What you WROTE is wrong because it means
    [tex] a 1 - \frac{r^{n+1}}{1} - r,[/tex]
    but if it was written properly as a(1-rn+1)/(1-r), it would be correct. What makes you think it is wrong?

    RGV
     
  11. Nov 13, 2012 #10

    SammyS

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    Write out a few terms:

    [itex]\displaystyle \ 2S=2\sum_{i=0}^{n}\left(2\right)^{i}\ [/itex]
    [itex]\displaystyle =2\left(2^0+2^1+2^2+2^3+\dots+2^{n-1}+2^{n}\right)[/itex]

    [itex]\displaystyle =2^1+2^2+2^3+2^4+\dots+2^{n}+2^{n+1}[/itex]

    [itex]\displaystyle =\sum_{i=1}^{n}\left(2\right)^{i}\ \ +\ \ 2^{n+1}\ [/itex]
     
  12. Nov 14, 2012 #11
    How right you are, I was over complicating it in my head, thanks.

    I was unnecessarily trying to minipulate it and ended up putting in the wrong value, like a circle through a square hole. Rather than just flipping it all.
    Yeah, I'll not neglect the brackets again.

    Legend, well explained, thanks!
     
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