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Math Help - binomial

  1. #1
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    binomial

    How would you find the term x^2*y*z^5 in the expansion of (x-2y+z)^8? First you have to expand ((x-2y)+z)^8.

    I understand how to find the coefficient but I don't really understand how to find the term.
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  2. #2
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    Quote Originally Posted by JQ2009 View Post
    How would you find the term x^2*y*z^5 in the expansion of (x-2y+z)^8? First you have to expand ((x-2y)+z)^8.

    I understand how to find the coefficient but I don't really understand how to find the term.
    Try to do it in two steps: as (a+b)^n=\sum\limits_{k=0}^n\binom {n}{k}a^kb^{n-k}, we get:

    ((x-2y)+z)^8=\sum\limits_{k=0}^n\binom {8}{k}(x-2y)^kz^{8-k}

    So you must choose k = 3 above to get z^5:  \binom {8}{3}(x-2y)^3z^5 , and now you must choose the correct power in (x-2y)^3 to finally get x^2yz^5

    Tonio
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  3. #3
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    What do you mean with choose the correct power?
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