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Math Help - Finding the next terms in series

  1. #1
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    Finding the next terms in series

    Given the geometric Series:
    1/5 + 1/25 + 1/125 + .........


    (a) Find the next two terms.

    (b) Determine if it is possible to find
    s.


    (c) If possible, find
    sand then express this sum using the sigma notation.

    Any help guys? I can understand that the next term will be 1/625 and then 1/3125? but I can't understand b and c.

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  2. #2
    Member Glaysher's Avatar
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    Quote Originally Posted by bobchiba View Post
    Given the geometric Series:
    1/5 + 1/25 + 1/125 + .........


    (a) Find the next two terms.

    (b) Determine if it is possible to find
    s.


    (c) If possible, find
    sand then express this sum using the sigma notation.

    Any help guys? I can understand that the next term will be 1/625 and then 1/3125? but I can't understand b and c.

    b) Only possible if -1 < r < 1. In this case r = 1/5 so yes (r is common ratio)
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bobchiba View Post
    Given the geometric Series:
    1/5 + 1/25 + 1/125 + .........


    (a) Find the next two terms.


    A geometric sequence a sequence of the form:
    a_n = a*r^(n - 1) for n = 1,2,3,4,5... where a_n is the nth term, a is the first term, r is the common ratio, and n is the current number of the term.

    here we have a = ar^0 = 1/5
    and we have a_2 = ar = 1/25

    => a_2/a = ar/ar^0 = r = (1/25)/(1/5) = 1/5

    so our sequence is: a_n = (1/5)(1/5)^(n - 1) for n = 1,2,3,4,5...

    so the next two terms are a_4 and a_5

    a_4 = (1/5)(1/5)^(4 - 1) = (1/5)^4 = 1/625
    a_5 = (1/5)(1/5)^(5 - 1) = (1/5)^5 = 1/3125

    any questions?


    (b) Determine if it is possible to find
    s.


    since |r|<1 it is possible to find S_infinity


    (c) If possible, find
    sand then express this sum using the sigma notation.

    Any help guys? I can understand that the next term will be 1/625 and then 1/3125? but I can't understand b and c.

    For a geometric sequence where |r|<1, the sum to infinity of all the terms is given by:

    S_infinity = a/(1 - r) = (1/5)/(1 - (1/5)) = (1/5)/(4/5) = 1/4

    express this in sigma notation we have:

    SUM{n=0 to infinity}ar^n = SUM{n=0 to infinity}(1/5)(1/5)^n = SUM{n=0 to infinity}(1/5)^(n+1)

    the sum should mean the sumation sign, write n = 0 at the bottom and the infinity symbol at the top
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