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Sum to Infinity of a Geometric Series

  1. Aug 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Q.: A geometric series has first term a and common ratio r. Its sum to infinity is 12. The sum to infinity of the squares of the terms of this geometric series is 48. Find the values of a and r.

    Ans.: From text book: a = 6, r = 1/ 2


    2. Relevant equations

    Eq.: S[itex]\infty[/itex] = [itex]\frac{a}{1 - r}[/itex]

    3. The attempt at a solution

    S[itex]\infty[/itex] = [itex]\frac{a}{1 - r}[/itex] = 12
    a = 12 ( 1 - r)
    a = 12 - 12r
    a = 1 - r

    S[itex]\infty[/itex] = [itex]\frac{a^2}{1 - r^2}[/itex]

    [itex]\frac{a^2}{1- (a^2r^2/ a^2)}[/itex] = 48
    [itex]\frac{r^2 - 2r + 1}{1 - (r^2((1 -r)^2))/ ((1 - r)^2)}[/itex] = 48

    [itex]\frac{r^2 - 2r + 1}{1 - ((r^2 - r^3)^2)/ (r^2 - 2r + 1)}[/itex] = 48

    [itex]\frac{r^2 - 2r + 1}{1 - (r^6 - 2r^5 + r^4)/ (r^2 -2r + 1}[/itex] = 48

    [itex]\frac{r^2 - 2r + 1}{1 - r^4 - 2r^4 + r^4}[/itex] = 48

    [itex]\frac{r^2 - 2r + 1}{1}[/itex] = 48
    r^2 - 2r + 1 -48
    r^2 - 2r + 1 -47
    (r - 47)(r - 1) ...?

    from answer above, though, r should = 1/ 2.

    Can anyone help me spot where I have gone wrong? Thank you.
     
    Last edited: Aug 11, 2011
  2. jcsd
  3. Aug 11, 2011 #2

    HallsofIvy

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    Science Advisor

    Okay, if "S" is a geometric series:
    [tex]\sum_{n=0}^\infty ar^n= \frac{a}{1- r}[/tex]
    with a and r, then the sum of squares,
    [tex]\sum_{n=0}^\infty a^2 (r^n)^2= \sum_{n=0}^\infty a^2(r^2)^n[/tex]
    is also a geometric series, with [itex]a^2[/itex] and [itex]r^2[/itex]
    So, as you say,
    [tex]\frac{a}{1- r}= 12[/tex]
    and
    [tex]\frac{a^2}{1- r^2}= 48[/tex]

    To solve that, I recommend dividing the second equation by the square of the first. (Recall that [itex]1- r^2= (1- r)(1+ r)[/itex].)
     
  4. Aug 11, 2011 #3
    Hi odolwa99

    I think there is a much easier way to solve the problem

    Like you know the s infinity sum of a geometric serie of ratio <1 is a0(1-q) where a0 is the first term and q is the ratio

    In our case the infinity sum is a/(1-r) = 12 -> a = 12(1-r)

    Now see: The sum of the squares is also a geometric serie, first term a² and ratio r², so the sum is

    a²/(1-r²) = 48 -> a² = 48(1-r²)

    So 144(1-r)² = 48(1-r²) -> 3(1-r)=(1+r) -> r = 1/2
     
  5. Aug 11, 2011 #4
    Ah, fantastic. Thank you very much. Btw, I'm getting used to latex so thats why the question has changed apppearance, and thanks to everyone for your help.
     
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