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Math Help - Prove this .

  1. #1
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    Prove this .

    Prove that \sum^{n}_{r=1}(r+1)2^{r-1}=n2^n .
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Prove that \sum^{n}_{r=1}(r+1)2^{r-1}=n2^n .
    Telescoping series:

    (r+1)2^{r-1}=r2^r-(r-1)2^{r-1}

    CB
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  3. #3
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    Thanks Captain , but how did you get that expression ? Is it by expansion ?
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    Quote Originally Posted by thereddevils View Post
    Thanks Captain , but how did you get that expression ? Is it by expansion ?
    First I deduce what method is appropriate for high school level, which made me suspect a telescoping series, then noting that 0\times 2^0=0 is a give away for writting the general term as r2^r-(r-1)2^{r-1}, since then the only part of the last term of the series that does not cancel is n2^2, and from the first term is 0.

    CB
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  5. #5
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    I know what a telescoping series is but i just don understand how you get this :

     <br />
(r+1)2^{r-1}=r2^r-(r-1)2^{r-1}<br />
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  6. #6
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    Quote Originally Posted by thereddevils View Post
    I know what a telescoping series is but i just don understand how you get this :

     <br />
(r+1)2^{r-1}=r2^r-(r-1)2^{r-1}<br />
    r 2^{r-1} + 2^{r-1} = r 2^{r-1} (2 - 1) + 2^{r-1} = r 2^r - r 2^{r - 1} + 2^{r-1} = r 2^r - 2^{r-1} (r - 1).
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