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Math Help - Series question

  1. #1
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    Series question

    Hi
    The question given is

    Prove by mathematical induction that

    \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{n}{3^n} = \frac{1}{4}(3-\frac{2n + 3}{3^n})

    I've spent quite a bit of time on this, playing around with it, but can't seem to make any head-way. If anyone can give any hints or tips at all i'd be rather grateful.

    -Dave
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  2. #2
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    Quote Originally Posted by walleye View Post
    Hi
    The question given is

    Prove by mathematical induction that

    \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{n}{3^n} = \frac{1}{4}(3-\frac{2n + 3}{3^n})

    I've spent quite a bit of time on this, playing around with it, but can't seem to make any head-way. If anyone can give any hints or tips at all i'd be rather grateful.

    -Dave

    For n = 1 we get:

    Left side: \frac{1}{3}

    Right side: \frac{1}{4}\left(3-\frac{2\cdot 1+3}{3^1}\right)=\frac{1}{4}\left(3-\frac{5}{3}\right)=\frac{1}{4}\left(\frac{4}{3}\ri  ght)=\frac{1}{3}...check.

    Assume now the claim's true for n and let's try to show for n+1:

    \frac{1}{3}+\frac{2}{3^2}+\ldots+\frac{n}{3^n}+\fr  ac{n+1}{3^{n+1}}=\frac{1}{4}\left(3-\frac{2n+3}{3^n}\right)+\frac{n+1}{3^{n+1}} =\frac{1}{4}\left(\frac{3^{n+1}-3-2n}{3^n}+\frac{4n+4}{3^{n+1}}\right) = \frac{1}{4}\left(\frac{3^{n+2}-9-6n+4n+4}{3^{n+1}}\right) .

    Well, now with a little (very little, indeed) algebra check that the rightmost side equals the formula with n+1 instead n, and we're done.

    Tonio
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  3. #3
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    I'm sorry, I forgot to thank you for this.

    after a bit of reworking I finally got it.

    Thanks
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  4. #4
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    Tonio, sorry, doesn't your last expression evaluate to \frac{1}{4}\left(3-\frac{2n+\color{red}5}{3^{n+1}}\right) ?
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  5. #5
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    Quote Originally Posted by scorpion007 View Post
    Tonio, sorry, doesn't your last expression evaluate to \frac{1}{4}\left(3-\frac{2n+\color{red}5}{3^{n+1}}\right) ?


    Indeed it does...just as needed.

    Tonio
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  6. #6
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    But don't we need

    <br />
\frac{1}{4}\left(3-\frac{2n+3}{3^{n+1}}\right)<br />
    ?

    Ah, nevermind! I forgot to substitute n+1 in 2(n+1)
    Last edited by scorpion007; March 20th 2010 at 07:44 PM.
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  7. #7
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    You can also differentiate 1+x+x^2+\dots+x^n = \frac{1-x^{n+1}}{1-x} with respect to x, multiply both sides by x and set x=\frac{1}{3}.
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