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Sum of geometric series

  1. Dec 13, 2009 #1
    I have to find the sum of [tex]\sum9(2/3)^n[/tex] and I get a/1-r where a=9 and r=2/3.....but I know a=6 and not 9. Can someone point out to me what I am doing wrong? The sum is from n=1 to infinity.


    EDIT: I am thinking I take a(1) which is 6 as the a in a/(1-r), is this correct?
    Last edited: Dec 13, 2009
  2. jcsd
  3. Dec 14, 2009 #2
  4. Dec 14, 2009 #3


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    You left out important information! What is the beginning index for your sum?

    [tex]\sum_{n=0}^\infty a r^n= \frac{a}{1- r}[/itex]

    [tex]\sum_{n=1}^\infty a r^n= r \sum_{n=0}^\infty a r^n[/tex]
    where I have factored out an "r" to reduce the first index to 0,
    [tex]\sum_{n=1}^\infty a r^n= \left(\sum_{n=0}^\infty a r^n\right) - a[/tex]
    where I have subracted off the [itex]ar^0[/itex] term.

    In particular, with a= 9, r= 2/3,
    [tex]\sum_{n=1}^\infty 9 (2/3)^n= (2/3)\sum_{n=0}^\infty 9 (2/3)^n[/tex]
    [tex]= \frac{2}{3}\frac{9}{1- 2/3}= \frac{6}{1- 2/3}= 18[/tex]

    [tex]\sum_{n=1}^\infty 9(2/3)^n= \left(\sum_{n=0}^\infty 9 (2/3)^n\right)- 9[/tex]
    [tex]= \frac{9}{1- 2/3}- 9= 27- 9= 18[/tex]
  5. Dec 14, 2009 #4
    \sum_{n=1}^\infty a r^n= r \sum_{n=0}^\infty a r^n

    hmm, may i know how to factorise that r please?
  6. Dec 14, 2009 #5


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    It's just the "distributive law": [itex]ar+ ar^2+ ar^3+ \cdot\cdot\cdot= r(a+ ar+ ar^2+ \cdot\cdot\cdot[/itex]
  7. Dec 14, 2009 #6
    owh, thank you very much
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