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Geometric Series Question

  1. Apr 14, 2013 #1
    [itex]\sum^{\infty}_{k=4}[/itex] [itex]\frac{1}{5^{k}}[/itex]


    ar^n=a/(1-r)


    Here is what I did:

    a=5
    r=1/625

    (1/5)/(1-1/625)=(1/5)/(624/625)=625/3120=125/624

    The Answer is 1/500

    Where am I going wrong?
     
  2. jcsd
  3. Apr 14, 2013 #2

    Dick

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    You've got a and r wrong. a is the first term in the series and r is the common ratio. Try again.
     
  4. Apr 14, 2013 #3

    Fredrik

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    Can you rewrite the sum in the form
    $$\frac{1}{5^4}\sum_{i=0}^\infty(\text{something})_i?$$ This isn't absolutely necessary, but if you do, you can use the version of the formula with a=1.
     
  5. Apr 14, 2013 #4
    i meant a to be 1/5...
    the first term was 1/625
     
  6. Apr 14, 2013 #5
    If I do it that way,r would equal 1/5 right?

    $$\frac{1}{5^4}\sum_{i=0}^\infty(\text{(1/5)^n})?$$
     
  7. Apr 14, 2013 #6

    Fredrik

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    Yes, assuming that you meant what I hope you meant when you wrote ar^n=a/(1-r), which doesn't really make sense. :smile:
     
  8. Apr 14, 2013 #7
    It worked, thank you so much
     
  9. Apr 14, 2013 #8
    The sum of an infinite geometric series of the form ##\displaystyle\sum_{n=0}^\infty ar^n=\dfrac{a}{1-r}##, for ##|r|<1##.

    Your given series starts off with ##n=4##. Think you can rewrite the series to make it start with ##n=0##?
     
  10. Apr 14, 2013 #9
    I didn't know in order for the formula to work I had to make n=0. Thanks
     
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