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anonymous

  • one year ago

What do I do, I am unsure how to interpret it as a geometric series. http://i.imgur.com/pyZd2ip.png Thanks

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  1. Empty
    • one year ago
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    You can make a substitution to make it easier perhaps? \[x=(1+c)^{-1}\] \[\sum_{n=1}^\infty x^n =8\]

  2. jtvatsim
    • one year ago
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    I agree with @empty. You can then use the infinite series formula for geometric series to get a value for x. Then solve for c after. That is, 1/(1-x) = 8 solve for x. You will get x = something. Then, x = 1/(1+c) so, something = 1/(1+c) solve for c.

  3. anonymous
    • one year ago
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    Ah yes thanks guys!

  4. anonymous
    • one year ago
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    Let \(x = 1+c\) Partial sum is\[s_n = x^{-1} + x^{-2} + x^{=3} + ... + x^{-n}\]Multiply both sides by \(x\)\[xs_n = 1 + x^{-1} + x^{-2} + ... + x^{-n+1}\]Subtract 1st equation from 2nd equation\[(x-1)s^n = 1 + x^{-n}\]\[s_n = \frac{ 1 + x^{-n} }{ x-1 }\]Sum of infinite series is\[\lim_{n \rightarrow \infty} s_n = \lim_{n \rightarrow \infty} \frac{ 1 + x^{-n} }{ x-1 } = 8\]\[\frac{ 1 }{ x-1 } = 8\]Solve for \(x\) and then for \(c\).

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