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Thread: From a series how do we get this result?

  1. #1
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    From a series how do we get this result?

    :
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  2. #2
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    Re: From a series how do we get this result?

    Quote Originally Posted by math951 View Post
    :
    This all I can offer you:
    $\displaystyle \sum\limits_{k = 1}^x {{{\left( {\frac{5}{6}} \right)}^{k - 1}}} = \frac{{1 - {{\left( {\tfrac{5}{6}} \right)}^x}}}{{1 - \left( {\tfrac{5}{6}} \right)}}$

    Because it $|r|<1$ then $\displaystyle \sum\limits_{k = 1}^x {{r^{k - 1}}} = \frac{{1 - {r^x}}}{{1 - r}}$
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  3. #3
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    Re: From a series how do we get this result?

    If you want a little more detail, let $\displaystyle S(n)= 1+ r+ r^2+ \cdot\cdot\cdot+ r^n$. $\displaystyle S(n)= 1+ (r+ r^2+ \cdot\cdot\cdot+ r^n)= 1+ r(1+ r+ \cdot\cdot\cdot+ r^{n-1})$.


    Add and subtract $\displaystyle r^n$ inside the parentheses:
    $\displaystyle S(n)= 1+ r(1+ r+ \cdot\cdot\cdot+ r^{n-1}+ r^n- r^n)= 1+ r(1+ \cdot\cdot\cdot+ r^n)- r^{n+1}$
    $\displaystyle S(n)= 1+ rS(n)- r^{n+1}$
    $\displaystyle S(n)- rS(n)= (1- r)S(n)= 1- r^{n+1}$
    $\displaystyle S(n)= \frac{1- r^{n+1}}{1- r}$.

    Here $\displaystyle r= \frac{5}{6}$.
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