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Math Help - Need help simplifying/factoring

  1. #1
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    Need help simplifying/factoring

    This problem is for a discrete math class, but requires factoring/simplifying. I haven't attempted anything like this in about 10 years.

    I need  (\frac{(k(k+1))}{2})^2 + (k+1)^3

    to look like this:  (\frac{(k+1)(k+2)}{2})^2

    If that's possible. This may be asking alot, but a brief explanation of the simplification would be fantastic.

    Thank you.
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  2. #2
    Junior Member
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    Quote Originally Posted by webhound2 View Post
    This problem is for a discrete math class, but requires factoring/simplifying. I haven't attempted anything like this in about 10 years.

    I need  (\frac{(k(k+1))}{2})^2 + (k+1)^3

    to look like this:  (\frac{(k+1)(k+2)}{2})^2

    If that's possible. This may be asking alot, but a brief explanation of the simplification would be fantastic.

    Thank you.
     (\frac{(k(k+1))}{2})^2 + (k+1)^3

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

    Now you can see that (k+1)^2 is a common factor, expression becomes:

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

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

     =(\frac{(k+1)(k+2)}{2})^2
    Last edited by Ithaka; February 5th 2011 at 08:54 PM. Reason: LATEX error
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  3. #3
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    that's perfect. It seems so simple now. Thank you.
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  4. #4
    MHF Contributor
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    Thanks
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    To give an explanation....

    You want to get

    \displaystyle\left[\frac{k(k+1)}{2}\right]^2+(k+1)^3

    to

    \displaystyle\left[\frac{(k+1)(k+2)}{2}\right]^2

    (k+1)^2 is common and you want 2^2 in the denominator

    hence express \displaystyle\ (k+1)^3=\frac{4(k+1)(k+1)^2}{4}=4(k+1)\left[\frac{k+1}{2}\right]^2

    Hence we get

    \displaystyle\ k^2\left[\frac{k+1}{2}\right]^2+4(k+1)\left[\frac{k+1}{2}\right]^2=\left(k^2+4k+4\right)\left[\frac{k+1}{2}\right]^2

    =\displaystyle\ (k+2)^2\left[\frac{k+1}{2}\right]^2=\left[\frac{(k+1)(k+2)}{2}\right]^2
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