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Math Help - Geometric series.

  1. #1
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    Geometric series.

    \sum^n _{k=1} \pi^k-3 =[\pi^n-1][\pi-1]^{-1}-3n correct?

    The back of my book is saying: \sum^n _{k=1} \pi^k-3 =\pi[\pi^n-1][\pi-1]^{-1}-3n however, I am not really seeing where the pi is coming from. Suggestions?
    Last edited by mr fantastic; January 6th 2011 at 04:53 PM. Reason: Re-titled.
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    Quote Originally Posted by quantoembryo View Post
    \sum^n _{k=1} \pi^k-3 =[\pi^n-1][\pi-1]^{-1}-3n correct?

    The back of my book is saying: sum^n _{k=1} \pi^k-3 =\pi[\pi^n-1][\pi-1]^{-1}-3n however, I am not really seeing where the pi is coming from. Suggestions?
    Dear quantoembryo,

    It is because the summation is takan from k=1.

    \displaystyle\sum_{k=0}^{k=n}r^k=\frac{1-r^n}{1-r}

    But, \displaystyle\sum_{k=1}^{k=n}r^k=\frac{r(1-r^n)}{1-r}
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  3. #3
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    Quote Originally Posted by quantoembryo View Post
    \sum^n _{k=1} \pi^k-3 =[\pi^n-1][\pi-1]^{-1}-3n correct?

    The back of my book is saying: \sum^n _{k=1} \pi^k-3 =\pi[\pi^n-1][\pi-1]^{-1}-3n however, I am not really seeing where the pi is coming from. Suggestions?
    You are using an incorrect formula for the sum of a geometric series. I suggest you go back and carefully review the correct formula (and also paying attention to the fact that you're sumiing from k = 1 not k = 0).
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    Thanks for the help, I figured it out
    Last edited by mr fantastic; January 6th 2011 at 05:30 PM.
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