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Math Help - Binomial Expansions with negative and fractional Expontents

  1. #1
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    Binomial Expansions with negative and fractional Expontents

    I know, I know two threads in 5 minutes, but I can't get this one!

    Binomial expansions are all fine and dandy, but when the exponent is a negative, a fraction, or both I am a little stumped.

    Write the fourth term in the binomial expansion for:
    (a - b)^-(1/5)

    and I don't even know where to begin!
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    Quote Originally Posted by Crell View Post
    I know, I know two threads in 5 minutes, but I can't get this one!

    Binomial expansions are all fine and dandy, but when the exponent is a negative, a fraction, or both I am a little stumped.

    Write the fourth term in the binomial expansion for:
    (a - b)^-(1/5)

    and I don't even know where to begin!
    Are you familiar with the binomial theorem?

    Binomial theorem - Wikipedia, the free encyclopedia
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  3. #3
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    Quote Originally Posted by adkinsjr View Post
    Are you familiar with the binomial theorem?

    Binomial theorem - Wikipedia, the free encyclopedia

    Yes, but I don't know how to apply it with negative, and fractional exponents.

    Binomial expansions are all fine and dandy, but when the exponent is a negative, a fraction, or both I am a little stumped.
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    Quote Originally Posted by Crell View Post
    Yes, but I don't know how to apply it with negative, and fractional exponents.
    Ok, I think it's the same basic idea, you would just have an infinite number of terms. It's not easy to do.

    http://www.mathhelpforum.com/math-he...exponents.html
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  5. #5
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    Quote Originally Posted by Crell View Post
    I know, I know two threads in 5 minutes, but I can't get this one!

    Binomial expansions are all fine and dandy, but when the exponent is a negative, a fraction, or both I am a little stumped.

    Write the fourth term in the binomial expansion for:
    (a - b)^-(1/5)

    and I don't even know where to begin!
    If a \neq 0 I would divide by a

    \left[a\left(1-\frac{b}{a} \right)\right]^{-\frac{1}{5}} = a^{-\frac{1}{5}} \times \left(1-\frac{b}{a}\right)^{-\frac{1}{5}}

    The first part is easy enough and the second part can be expressed by manipulating

    (1+x)^n = 1+nx+\frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ... + \frac{n!}{n!}x^n
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