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Math Help - Stuck on an Induction Problem

  1. #1
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    Question Stuck on an Induction Problem

    Prove that:
    \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + ... + \frac{n}{3^n} = \frac{3}{4} - \frac{2n+3}{4*3^n}

    My work so far:
    Let P(n) denote the statement that \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + ... + \frac{n}{3^n} = \frac{3}{4} - \frac{2n+3}{4*3^n}

    P(1) = \frac{1}{3^1} = \frac{3}{4} - \frac{5}{4*3^1}
     \frac{1}{3} = \frac{1}{3}
    The statement holds true for the case n = 1.

    Assume that P(n) is true for when n is equal to a natural constant k,
    P(k) = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + ... + \frac{k}{3^k} = \frac{3}{4} - \frac{2k+3}{4*3^k}

    Prove that P(n) is true when n = k + 1
    P(k+1) = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + ... + \frac{k}{3^k} + \frac{k+1}{3^{k+1}}= \frac{3}{4} - \frac{2(k+1)+3}{4*3^{k+1}}

    Replacing the terms on the left hand side of P(k+1) with the right hand side of P(k) I then had:

    P(k+1) = \frac{3}{4} - \frac{2k+3}{4*3^k} + \frac{k+1}{3^{k+1}} = \frac{3}{4} - \frac{2(k+1)+3}{4*3^{k+1}}

    This is as far as I could get! I need assistance simplifying the left hand side of this last line, assuming I have worked this through correctly. I have not worked too much with sequences so my trouble is mainly with getting a common denominator. Thanks for anyone that helps with this, I know it is a lot to read through.
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  2. #2
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    The expression can be simplified by finding the lowest common denominator of 4\cdot 3^k and 3^{k+1}.
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  3. #3
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    That's what I was having problems with, is the lowest common denominator then 4*3^{k+1}? And I need to take 3^{k+1} = (3^k)(3^1) ?
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  4. #4
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    The lowest comon denominator is OK!

    If you make the calculations carefully, you will have to compare

    -((2k+1)+3) with -3(2k+3)+4(k+1)
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