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Math Help - Binomial expansion

  1. #1
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    Binomial expansion

    Hello, i am unsure how to do this question... Find the value of the term in the expansion of (1-1/x)^8 which is independent of x. How do i do this? I have no problem with normal expansions but how do you do one with a fraction in?

    Thankyou,
    Kris.
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  2. #2
    Eater of Worlds
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    I assume independent of x means it has no x term and is just a constant.

    Then it's a matter of observation. It's 1. Becasue of the 1 in (1-\frac{1}{x})^{8}

    If you have (p+q)^{8}, just sub in -1/x where the q goes in the expansion. Fractions are no different than an integer.

    1^{8}+8(\frac{-1}{x})+28(\frac{-1}{x})^{2}+...........+(\frac{-1}{x})^{8}
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  3. #3
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    Ok thankyou, yes i am not really sure what is meant by 'independent of x' as the expresion is (x-1/x)^8 is this just the constant in the expansion?

    Thankyou for your help,
    Kris
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  4. #4
    Senior Member JaneBennet's Avatar
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    I suspect there is a typo in your original post: you probably want the constant term in the expansion of \left(\color{red}x\color{black}-\frac{1}{x}\right)^8. This would be the term ^8\mathrm{C}_4\,x^4\,\left(-\frac{1}{x}\right)^4.
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