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Homework Help: G.P. = [itex] \sum_{n=1}^k a_{1}(\frac{a_{n+1}}{a_{n}})^{n-1} [/itex]

  1. Jun 3, 2013 #1
    A geometric series is [itex]S_{n}= a_{1} + a_{2} + a_{3} + ...+ a_{n} = a_{1}r^{0} + a_{2}r^{1} + a_{3}r^{2} + ...+ a_{n}r^{n-1}[/itex]

    In a geometric series, the first term = [itex]a_{1}[/itex]

    The common ratio = [itex]\frac{a_{n+1}}{a_{n}}[/itex]

    The nth term of a geometric sequence is [itex]a_{n}= a_{1}r^{n-1}= a_{1}(\frac{a_{n+1}}{a_{n}})^{n-1}[/itex]

    The question is:

    In the summand of the sigma notation, can the general term of geometric sequence formula be expressed this way:

    [tex]S_{n}= \sum_{i=1}^n a_{1}r^{i-1} = \sum_{i=1}^n a_{1}(\frac{a_{i+1}}{a_{i}})^{i-1}[/tex]
  2. jcsd
  3. Jun 3, 2013 #2


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    It can, but why would you want to? It loses the information that the ratio is a constant.
  4. Jun 3, 2013 #3


    Staff: Mentor

    The two expressions you show aren't equal unless r = 1.

    You should write your (finite) geometric series like so:
    ##S_{n}= ar^{0} + ar^{1} + ar^{2} + ...+ ar^{n-1} ##

    There is no need for subscripts in a geometric series. The first term is a, the next is ar, the one after that is ar2, and so on.
    The common ratio is r. Each term in the series is r times the previous term.
    The n-th term is arn-1.
    It's much simpler like this:
    $$S_n = \sum_{k = 0}^{n-1} ar^k$$
    Last edited: Jun 3, 2013
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