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Math Help - binomial theorem help

  1. #1
    Newbie
    Joined
    Feb 2009
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    Unhappy binomial theorem help

    hey, I'm really struggling on this question i just cant seem to find the right method to get to the answer which is 0.4375

    find and simplify the term independent of x in the expansion

    <br />
({\frac{1}{2x}}+x^3)^8<br />


    thanks xx
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  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
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    South Coast of England
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    Binomial Coefficients

    Hello alyson
    Quote Originally Posted by alyson View Post
    hey, I'm really struggling on this question i just cant seem to find the right method to get to the answer which is 0.4375

    find and simplify the term independent of x in the expansion

    <br />
({\frac{1}{2x}}+x^3)^8<br />


    thanks xx
    \left({\frac{1}{2x}}+x^3\right)^8 = \left(\frac{1+2x^4}{2x}\right)^8

    = \left(\frac{1}{2x}\right)^8(1+2x^4)^8

    You now need to find the term in x^8 in the expansion of (1+2x^4)^8, to cancel the \frac{1}{x^8} outside. This term is:

    \binom{8}{2}(2x^4)^2

    So the coefficient is:

    \frac{1}{2^8}\times \frac{8\times 7 \times 2^4}{2} = \frac{7}{16} = 0.4375

    Grandad
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