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Math Help - coefficient

  1. #1
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    coefficient

    in the expansion of
    wath is the coefficient of
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  2. #2
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    Hello, dapore!

    \text{In the expansion of: }\:\left( x-\sqrt{x}-1 \right)^{7},\;\text{ what is the coefficient of }\,x^{\frac{5}{2}}\:?

    \text{The only terms containing }\,x^{\frac{5}{2}}\,\text{ are: }\;\left(x\right)^2\!\left(\text{-}x^{\frac{1}{2}}\right)\!\left(\text{-}1\right)^4,\;\;\left(x\right)^1\!\left(\text{-}x^{\frac{1}{2}}\right)^3\!\left(\text{-}1\right)^3, . \left(x\right)^0\!\left(x^{\frac{1}{2}}\right)^5\!  \left(\text{-}1\right)^2


    We have: . {7\choose2,1,4}x^2\left(\text{-}x^{\frac{1}{2}}\right)(\text{-}1)^4 \;=\;105\left(-x^{\frac{5}{2}}\right)

    . . . . . . . . {7\choose1,3,3}x\left(\text{-}x^{\frac{1}{2}}\right)^3(\text{-}1)^3 \;=\;140\left(+x^{\frac{5}{2}}\right)

    . . . . . . . . {7\choose0,5,2}x^0\left(\text{-}x^{\frac{1}{2}}\right)^5(\text{-}1)^2 \;=\; 21\left(-x^{\frac{5}{2}}\right)


    Therefore: . -105x^{\frac{5}{2}} + 140x^{\frac{5}{2}} - 21x^{\frac{5}{2}} \;\;=\;\; {\color{blue}14}\,x^{\frac{5}{2}}

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, dapore!


    \text{The only terms containing }\,x^{\frac{5}{2}}\,\text{ are: }\;\left(x\right)^2\!\left(\text{-}x^{\frac{1}{2}}\right)\!\left(\text{-}1\right)^4,\;\;\left(x\right)^1\!\left(\text{-}x^{\frac{1}{2}}\right)^3\!\left(\text{-}1\right)^3, . \left(x\right)^0\!\left(x^{\frac{1}{2}}\right)^5\!  \left(\text{-}1\right)^2


    We have: . {7\choose2,1,4}x^2\left(\text{-}x^{\frac{1}{2}}\right)(\text{-}1)^4 \;=\;105\left(-x^{\frac{5}{2}}\right)

    . . . . . . . . {7\choose1,3,3}x\left(\text{-}x^{\frac{1}{2}}\right)^3(\text{-}1)^3 \;=\;140\left(+x^{\frac{5}{2}}\right)

    . . . . . . . . {7\choose0,5,2}x^0\left(\text{-}x^{\frac{1}{2}}\right)^5(\text{-}1)^2 \;=\; 21\left(-x^{\frac{5}{2}}\right)


    Therefore: . -105x^{\frac{5}{2}} + 140x^{\frac{5}{2}} - 21x^{\frac{5}{2}} \;\;=\;\; {\color{blue}14}\,x^{\frac{5}{2}}
    Thank you, thank you, thank you
    But how we calculated

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  4. #4
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    Quote Originally Posted by dapore View Post
    Thank you, thank you, thank you
    But how we calculated

    That expression is equal to \frac{7!}{2!\,1!\,4!}

    I believe it comes from the coefficients of the binomial expansion
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