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Math Help - Series: Method of Difference

  1. #1
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    Series: Method of Difference

    Given S = (4/5) + (7/5^2) + (10/5^3) + ... + (31/5^10), express S in Σ form.
    By considering the difference between S and (1/5)S, find S.

    So this is how I did it:
    S = Σ (3r+1)/5^r (Upperlimit 10, Lower limit r = 1 )
    (1/5)S = Σ (3r+1)/5^(r+1)

    S - (1/5)S = Σ [(3r+1)/5^r - (3r+1)/5^(r+1)]
    = + 5/4 - 4/25 r = 1
    + 1/25 - 7/125 r = 2
    + 10/125 - 10/625 r = 3
    + ...
    I use the method of difference to solve but there don't seem to be any terms I can cancel away... Any idea what went wrong?
    THanks in advance!!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    In general \sum_{n=1}^{+\infty}(an+b)r^n=a\sum_{n=1}^{+\infty  }nr^n+b\sum_{n=1}^{+\infty}r^n. The second addend corresponds to a geometric series. For the first one and denoting H=\sum_{n=1}^{+\infty}nr^n

    \begin{Bmatrix} H=r+2r^2+3r^3+\ldots\\rH=r^2+2r^3+3r^4\ldots\end{m  atrix}\Rightarrow (1-r)H=r+r^2+r^3+\ldots

    Now you can easily find H
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  3. #3
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    No nee to cancel.

    S - (1/5)S = (4/5) + (7 - 4)/5^2 + (10 - 7)/5^3 + ............(31 - 28)/5^10 - 31/5^11

    Can you proceed now?
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