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Math Help - geometric progressions

  1. #1
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    geometric progressions

    show that the sum of n terms of the series

    \log a + \log (2a) + \log (4a) + \log (8a) + ....

    is \log [a^{n} \, 2^{n(n-1)/2}]
    Last edited by mr fantastic; August 12th 2009 at 04:41 AM. Reason: Improved latex and formatting
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  2. #2
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    Quote Originally Posted by Rose Wanjohi View Post
    show that the sum of n terms of the series

    \log a + \log (2a) + \log (4a) + \log (8a) + ....

    is \log [a^{n} \, 2^{n(n-1)/2}]
    Start by noting that the series can be written as

    \log a + (\log a + \log 2 ) + (\log a + 2 \log 2 ) + (\log a + 3 \log 2 ) + ....

    = n \log a + (1 + 2 + 3 + .... + n) \log 2 .

    The log rule \log A^B = B \log A wil also be useful.
    Last edited by CaptainBlack; August 13th 2009 at 07:41 AM.
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  3. #3
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    Hello, Rose!

    Show that: . \log a + \log (2a) + \log (4a) + \log (8a) + \hdots \log(2^{n-1}a) \;=\;\log \left[a^{n} \, 2^{\frac{n(n-1)}{2}}\right]

    \log a + \log(2a) + \log(4a) + \log(8a) + \hdots + \log(2^{n-1}a)

    . . = \;\log\bigg[a\cdot2a\cdot4a\cdot8a\cdots 2^{n-1}a\bigg]

    . . = \;\log\bigg[ ( 2\cdot4\cdot8  \cdots 2^{n-1}) (a\cdot a\cdot a\cdot a \cdots a)\bigg]

    . . = \;\log\bigg[\left(2\cdot2^2\cdot2^3 \cdots 2^{n-1}\right)(a^n)\bigg]

    . . = \;\log\bigg[ \left(2^{1+2+3+\hdots + (n-1)}\right)\left(a^n \right)\bigg]

    . . = \;\log\bigg[2^{\frac{n(n-1)}{2}}a^n\bigg]

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