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Math Help - Rearranging Algebra

  1. #1
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    Rearranging Algebra

    How do you rearrange from 2-\frac{k+2}{2^k}+\frac{k+1}{2^{k+1}} to give 2-\frac{k+3}{2^{k+1}}

    thank you!
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  2. #2
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    Hi BabyMilo!

    I think you have made a typing error!

    If you mean :

    2 - \frac{k + 2}{2^k} - \frac{k+1}{2^{k+1}}

    then you can arrange it to get

    2 - \frac{k+3}{2^{k+1}}

    let me know and if you are still stuck i'll type out the answer!
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  3. #3
    Super Member Anonymous1's Avatar
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    Lemme know if you need explanation for any particular step.

    2- \frac{k+2}{2^k} - \frac{k+1}{2^{k+1}} = 2- \frac{k+2}{2^k} - \frac{k+1}{{\color{red} 2\times2^k}} = 2- \frac{{\color{red} 2}(k+2)}{{\color{red} 2}\times2^k} - \frac{k+1}{2\times2^k} = 2- \frac{2(k+2) - (k+1)}{2\times2^k}= 2- \frac{{\color{red}2k+4 - k-1}}{2\times2^k}= 2 - \frac{k+3}{{\color{red}2^{k+1}}}
    Last edited by Anonymous1; March 19th 2010 at 12:07 PM.
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by BabyMilo View Post
    How do you rearrange from 2-\frac{k+2}{2^k}+\frac{k+1}{2^{k+1}} to give 2-\frac{k+3}{2^{k+1}}

    thank you!

    2-\frac{k+2}{2^k}+\frac{k+1}{2^(k+1)}

    = 2-\frac{k+2}{2^k}+\frac{k+1}{(2^k) 2}

    = 2-\frac{2(k+2)}{2^{(k+1)}}+\frac{k+1}{2^{(k+1)}}

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

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

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

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

    Sorry for the LaTex error on the first post
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  5. #5
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    Quote Originally Posted by harish21 View Post
    2-\frac{k+2}{2^k}+\frac{k+1}{2^(k+1)}

    = 2-\frac{k+2}{2^k}+\frac{k+1}{(2^k) 2}

    = 2-\frac{2(k+2)}{2^{(k+1)}}+\frac{k+1}{2^{(k+1)}}

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

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

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

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

    Sorry for the LaTex error on the first post
    the steps made it very clear.

    thanks
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