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Math Help - without expanding

  1. #1
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    without expanding

    without expanding, determine the ff

    term free of y in (2x^2 y^4 + x^3 y^-2) ^15

    tnx
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  2. #2
    MHF Contributor kalagota's Avatar
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    hi kabayan!

    find n such that y^{4(15-n)}y^{-2n}=y^0

    so basically, you only need to solve for 4(15-n)+(-2n)=0

    once you've solve this, then find \left(\begin{array}{r}15 \\ n \end{array}\right)(2x^2)^{15-n}(x^{3n})
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  3. #3
    Moo
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    A Cute Angle Moo's Avatar
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    Hello kalagota,

    Quote Originally Posted by kalagota View Post

    once you've solve this, then find \left(\begin{array}{r}15 \\ n \end{array}\right)(2x^2)^{15-n}(x^{3n})
    Instead of using an array, you can do {15 \choose n} [tex]{15 \choose n}[/tex] (there is another way but I don't remember)
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by Moo View Post
    (there is another way but I don't remember)
    [tex]\binom{15}{n}[/tex] gives \binom{15}{n}
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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Isomorphism View Post
    [tex]\binom{15}{n}[/tex] gives \binom{15}{n}
    this doesn't seem to work on my editor/compiler..
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