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Math Help - term of a series

  1. #1
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    term of a series

    The rth term of a series is given by

     <br />
u_r=(\frac{1}{3})^{3r-2}+(\frac{1}{3})^{3r-1}<br />

    Express \sum^{n}_{r=1}u_r in the form

    A(1-\frac{B}{27^n}) , where A and B are constants to be found ,
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    The rth term of a series is given by

     <br />
u_r=(\frac{1}{3})^{3r-2}+(\frac{1}{3})^{3r-1}<br />

    Express \sum^{n}_{r=1}u_r in the form

    A(1-\frac{B}{27^n}) , where A and B are constants to be found ,
     <br />
u_r=(\frac{1}{3})^{3r-1}\frac{3}{1}+(\frac{1}{3})^{3r-1}<br /> <br />

     <br />
 u_r=(\frac{1}{3})^{3r-1}(3+1)<br /> <br />

     <br />
  u_r=(\frac{1}{3})^{3r-1}(4) ~= 12\frac{1}{3^{3r}} ~= \frac{12}{27^r}<br /> <br />

    No need to tell you that this is a GP


    Sum = a\frac{(r^n-1)}{r-1}

    =\frac{4}{9} \frac{(1-\frac{1}{27^n})}{(1-\frac{1}{27})} <br /> <br />

    =\frac{4 \times 27}{9\times 26} {(1-\frac{1}{27^n})} <br /> <br />


    I think now you can continue to
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