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Algebra 2 - Geometric Series

posted by on .

I need help evaluating the geometric infinite series.

It's got the double zero thing over the E thing.

00
E(-1/2)^ n-1
(n=1)

  • Algebra 2 - Geometric Series - ,

    Substitute:

    n = k + 1

    Then k runs from zero to infinity

    The summand is (-1/2)^k

    So, the summation is:

    1/[1- (-1/2)] = 2/3

  • Algebra 2 - Geometric Series - ,

    How did you get k + 1?
    How would I do
    00
    E 3(0.4)^n-1
    n=1

    Would that be the same basic thing? Like 1/[1-3(0.4)]??

  • Algebra 2 - Geometric Series - ,

    You are free to shift your summation variable in any way you like. If you put n = k + 1 then what happens is that the lower limit of the summation of
    n = 1 corresponds to k = 0 while the upper limit remains infinity.

    In the summand you then replace n by
    k + 1 and you then see that the summation is now in the standard form of a sum from zero to infinity of a^k which is 1/(1-a)

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