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find the sum of the infinite series (2/9)^2-(2/9)^3+(2/9)^4-(2/9)^5+ ...

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asked by cecile
  1. (2/9)^2-(2/9)^3+(2/9)^4-(2/9)^5+
    This is like a geometric series with the first terms missing
    1 -1(2/9)^1 +1(2/9)^2-1(2/9)^3+(2/9)^4-(2/9)^5+
    so if we find the sum of that series and subtract [1 -1(2/9)^1], we have it
    so
    g = 1
    r = -2/9
    in Sn = g (1-r^n)/(1-r)
    n is infinite and |r| < 1
    so Sn =g/(1-r)
    Sn = 1/(1-2/9) = 1.286
    so we want
    1.286 -[1-2/9]
    .508

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    posted by Damon
  2. Whoa - typo
    Sn = 1/(1+2/9) because r = -2/9
    Sn = .818
    so we want
    .818 -[1-2/9]
    .040

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    2. πŸ‘Ž 0
    posted by Damon
  3. i need help

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    posted by ashlee and blake

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