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Math Help - Help with a proof.

  1. #1
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    Help with a proof.

    Hi, I'm stuck at proving the following question...

    Prove that for all n>0,

    1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n

    I've tried all sorts of different ways of solving this, but to no avail.

    Any help is appreciated
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by seven.j View Post
    Hi, I'm stuck at proving the following question...

    Prove that for all n>0,

    1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n

    I've tried all sorts of different ways of solving this, but to no avail.

    Any help is appreciated
    \sum_{k=1}^{n}\frac{k}{2^k}. Note that \sum_{k=1}<br />
^n x^k=\frac{x^{n+1}-x}{x-1}. Differentiating both sides and multiplying by x gives \sum_{k=1}^{n}k\cdot x^k=... figure the right side out.
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  3. #3
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    \sum\limits_{j=1}^{n}{\frac{j}{2^{j}}}=\sum\limits  _{j=1}^{n}{\sum\limits_{k=1}^{j}{\frac{1}{2^{j}}}}  =\sum\limits_{k=1}^{n}{\frac{1}{2^{k}}\left( \sum\limits_{j=0}^{n-k}{2^{-j}} \right)}=\frac{1}{2^{n}}\sum\limits_{k=1}^{n}{\fra  c{1}{2^{k}}\left( 2^{n+1}-2^{k} \right)}, you can do the rest, those are finite geometric sums.
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  4. #4
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    Hello, seven!

    Here's one way . . .


    Prove that for all n>0\!:

    . . \frac{1}{2}+ \frac{2}{2^2} + \frac{3}{2^3} +\:\hdots\:+ \frac{n}{2^n} \; =\; 2 - \frac{n+2}{2^n}
    \begin{array}{ccccc}<br />
\text{We have:} &S &=& \dfrac{1}{2} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \dfrac{4}{2^4} + \:\hdots\:+\dfrac{n}{2^n}\qquad\qquad \\ \\[-3mm]<br /> <br />
\text{Multiply by }\dfrac{1}{2}\!: & \dfrac{1}{2}S &=& \quad\;\; \dfrac{1}{2^2} + \dfrac{2}{2^3} + \dfrac{3}{2^4} + \hdots + \dfrac{n-1}{2^n} + \dfrac{n}{2^{n+1}}\end{array}

    . . . \text{Subtract: }\quad \frac{1}{2}S \;=\;\underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots + \frac{1}{2^n}}_{\text{geometric series}} - \frac{n}{2^{n+2}} .[1]

    The geometric series has the sum: . \frac{1}{2}\cdot\frac{1 - \left(\frac{1}{2}\right)^n}{1-\frac{1}{2}} \;=\;1 - \frac{1}{2^n}


    Then [1] becomes: . \frac{1}{2}S \;=\;\left(1 - \frac{1}{2^n}\right) - \frac{n}{2^{n+1}} \;=\;1 - \frac{n+2}{2^{n+1}}


    Multiply by 2: . S \;=\;2 - \frac{n+2}{2^n}

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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by seven.j View Post
    Hi, I'm stuck at proving the following question...

    Prove that for all n>0,

    1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n

    I've tried all sorts of different ways of solving this, but to no avail.

    Any help is appreciated
    This problem also can be done by induction pretty easily. if you're interested, you can try it that way. to me it was the most knee-jerk approach to try, and it worked out great. but you have lots of nice approaches here to choose from
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