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anonymous

  • 5 years ago

Please help answer this geometric question!!

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  1. anonymous
    • 5 years ago
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    For what values of x, \[x \neq-1\] will the following infinite geometric series have a finite sum? \[(x+1) + (x+2)^2 + (x+1)^3 ....\]

  2. dumbcow
    • 5 years ago
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    when -1 < x+1 < 0 --> -2 < x < -1 and 0< x+2 < 1 --> -2 < x < -1 this way the series converges and the sum will be finite

  3. anonymous
    • 5 years ago
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    pls explain properly

  4. anonymous
    • 5 years ago
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    no it's infinite 'cause the exponents are growing up

  5. dumbcow
    • 5 years ago
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    another way of looking at it is to separate it into 2 series and use the formula for sum of a geometric series Sn = a(1 - r^n)/(1-r) For series (x+1)^n , n is odd going to infinity, r = (x+1)^2 S = (x+1)(1-(x+1)^2n) / (1-(x+1)^2) = [x+1 -(x+1)^2n+1] / -(x^2+2x) For S to not be +-infinity, you need (x+1)^n to go to zero thus 0<|x+1| <1 S = (x+1)/-(x^2 +2x) For series (x+2)^n , n is even going to infinity, r = (x+2)^2 S = (x+2)^2(1-(x+2)^2n) / (1-(x+2)^2) = [(x+2)^2 -(x+1)^2n+2] / -(x^2 +4x+3) For S to not be +-infinity, you need (x+2)^n to go to zero thus 0<|x+2| <1 S = (x+2)^2/-(x^2 +4x +3) So Basically we need to find an x such that 0 < |x+1| < 1 and 0<|x+2| < 1 see prev post -2 < x < -1

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