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Math Help - quick fraction simplification help

  1. #1
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    Mar 2009
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    quick fraction simplification help

    Hey, just need some help with a fraction simplification problem.

    how can i make [k(k+1)(k+2)]/3 + (k+1)(k+2) = (k+1)(k+2)(k/3 + 1)

    and then take that step and have it equal [(k+1)(k+2)(k+3)]/3

    i slightly understand the second to third step but not how to go from the first to the second (factor out a k but how?) thanks.
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  2. #2
    MHF Contributor
    Joined
    Sep 2008
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    West Malaysia
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    Quote Originally Posted by Xender1 View Post
    Hey, just need some help with a fraction simplification problem.

    how can i make [k(k+1)(k+2)]/3 + (k+1)(k+2) = (k+1)(k+2)(k/3 + 1)

    and then take that step and have it equal [(k+1)(k+2)(k+3)]/3

    i slightly understand the second to third step but not how to go from the first to the second (factor out a k but how?) thanks.

    \frac{k(k+1)(k+2)}{3}+(k+1)(k+2)

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

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

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

     <br />
=(k+1)(k+2)(k/3 + 1)<br />
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