algebra

consider the infinite geometric series n=1 -4(1/3)^n-1 .

i need help with writing the first four terms of the series and finding the sum if it has a sum.

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  1. Do you mean
    a_n = 1 - 4(1/3)^(n-1)
    ?
    If so, then the sequence is
    1 - 4*1, 1 - 4/3, 1 - 4/9, 1 - 4/27 ...
    = -3, -1/3, 5/9, 23/27, ...
    But that's not a geometric sequence. I suspect you meant
    a = -4
    r = 1/3
    If so, then that sequence is
    -4, -4/3, -4/9, -4/27, ...
    And like all GP, S = 1/(1-r) = -4/(1 - 1/3) = -4/(2/3) = -4 * 3/2 = -6

    If I still didn't parse your text correctly, then I assume you can fix it and work it out using the methods above.

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    oobleck
  2. no, the problem that i have has Σ with an infinite sign and n = 1. after it it has -4(1/3)^n-1

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  3. then that is what I did in my 2nd half above.

    Σ -4(1/3)^(n-1) = -4 (1 + 1/3 + 1/9 + 1/27) ... = -6
    n=1

    I see I did make a typo: S = a/(1-r)
    not 1/(1-r)

    Looks like it's time for a review of your textbook.

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    oobleck

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