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Math Help - urgent please algebra solving...

  1. #1
    Junior Member
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    urgent please algebra solving...

    How can I go from

    ((n^2+(n+1)^2)/4) + 3 [(1/6)n(n+1)(2n+1)] + 2 [(1/2)n(n+1)]


    to this

    (1/4)n(n+1)(n+2)(n+3)
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  2. #2
    Junior Member eah1010's Avatar
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    Unhappy ummm...

    The only thing i got was

    5/2n^2 + 8n + 7/4
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  3. #3
    MHF Contributor
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    Quote Originally Posted by Natasha
    How can I go from
    ((n^2+(n+1)^2)/4) + 3 [(1/6)n(n+1)(2n+1)] + 2 [(1/2)n(n+1)]
    to this
    (1/4)n(n+1)(n+2)(n+3)
    Ok. So you have this equation:

    \frac{n^2+(n+1)^2}{4}+3\left(\frac{n(n+1)(2n+1)}{6  }\right)+2\left(\frac{n(n+1)}{2}\right)

    I would suggest getting a common denominator. Then expand things out, and simplify. There may be a more elegant way to do it, but it won't hurt to practice algebra.

    Jameson
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  4. #4
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    Right then...

    ((n^2+(n+1)^2)/4) + 3 [(1/6)n(n+1)(2n+1)] + 2 [(1/2)n(n+1)]

    After putting all the terms to a common denominator. Then expanding out, and simplifying the above I get

    ((12n^3+36n^2+14n+1)) / 12

    And I need to get

    (1/4)n(n+1)(n+2)(n+3)

    Can someone help the simple further steps to take thanks :-)
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  5. #5
    Junior Member
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    ((n^2+(n+1)^2)/4) + 3 [(1/6)n(n+1)(2n+1)] + 2 [(1/2)n(n+1)]

    After putting all the terms to a common denominator. Then expanding out, and simplifying the above I get

    ((4n^3+12n^2+8n+1)) / 4

    And I need to get

    (1/4)n(n+1)(n+2)(n+3)

    Can someone help the simple further steps to take thanks :-)
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Natasha
    How can I go from

    ((n^2+(n+1)^2)/4) + 3 [(1/6)n(n+1)(2n+1)] + 2 [(1/2)n(n+1)]


    to this

    (1/4)n(n+1)(n+2)(n+3)
    I suspect that you may have an typo in one or other
    of these expressions.

    Let n=0, then the first simplifies to 1/4,
    while the second simplies to 0.

    RonL
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  7. #7
    Junior Member
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    You are spot on. I had made a typo mistake... Resolve it now thanks
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