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What is the value of the summation of (2^n+1)/3^n?

  1. Apr 11, 2010 #1
    So the question asks: What is the value of the "summation of" 2n+1/3n from "n=1 to infinity."

    I changed 2n+1/3n into 2*(2/3)n so i could use it as a geometric series.

    So now i just use the rule "a/(1-r) = sum" where a = first term and r = ratio i get 2/(1-(2/3)) which = 6. The answer is 4, any suggestions?
     
  2. jcsd
  3. Apr 11, 2010 #2

    gabbagabbahey

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    Isn't your first term 2/3?
     
  4. Apr 11, 2010 #3
  5. Apr 11, 2010 #4
    Oh wait, so the first term is 4/3 so i get (4/3)/[1-(2/3)] = 4. Thank you guys!
     
  6. Apr 11, 2010 #5

    gabbagabbahey

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    The [itex]a[/itex] in the numerator takes care of that.

    [tex]\sum_{n=n_0}^\infty r^n=\frac{r^{n_0}}{1-r}=\frac{a}{1-r}[/tex]

    (When [itex]r<1[/itex] and [itex]n_0\geq0[/itex] of course)
     
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