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Math Help - Finding a Binomial Coefficient

  1. #1
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    Thumbs up Finding a Binomial Coefficient

    What is the coefficient of x^5 in the binomial expression (x - 2x^-2)^20 when multiplied by (1+3x^3)



    So far I have:
    (20 over k)(-1)^k (2)^k x^(20-3k)

    so then 20-3k = 5 such that k = 5

    So then (20 over 5 )(-1)^5 (2)^5

    But then I don't know how to handle multiplying it by ( 1 + 3x^3)
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  2. #2
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    Re: Finding a Binomial Coefficient

    (x - 2x^{-2})^{20}=a_{20}x^{20}+a_{17}x^{17}+\dots+a_5x^5+a_  2x^2+\dots+a_{-37}x^{37}+a_{-40}x^{-40} for some coefficients a_k. Also, 1+3x^3=b_0x^0+b_3x^3 for b_0=1, b_3=3. The coefficient of x^5 in (x - 2x^{-2})^{20}(1+3x^3) is \sum_{i+k=5}a_ib_k, i.e., a_2b_3+a_5b_0.
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  3. #3
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    Re: Finding a Binomial Coefficient

    Quote Originally Posted by ThatPinkSock View Post
    What is the coefficient of x^5 in the binomial expression (x - 2x^-2)^20 when multiplied by (1+3x^3)
    So far I have: (20 over k)(-1)^k (2)^k x^(20-3k)
    so then 20-3k = 5 such that k = 5
    So then (20 over 5 )(-1)^5 (2)^5
    but then I don't know how to handle multiplying it by ( 1 + 3x^3)
    I have a different take on this question from you and reply #2.
    Each term in \left(x-2x^{-2}\right)^{20} looks like \binom{20}{k}(-2)^{20-k}x^{3k-40}
    Now you want x^5 from [x^{3k-40}]~\&~3x^3{x^{3k-40}}
    So k=14~\&~k=15 WHY?
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