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Math Help - Constant term

  1. #1
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    Constant term

    How do I find the constant term in ((-x)^3+(-3/x))^24 ?
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  2. #2
    Behold, the power of SARDINES!
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    Binomial theorem.

    Quote Originally Posted by weasley74 View Post
    How do I find the constant term in ((-x)^3+(-3/x))^24 ?

    (a+b)^n=\sum_{k=0}^{n}\binom{n}{k}(a)^{n-k}b^{k}


    for yours we get

    ([(-x)]^3+(-3x^{-1}))^{24}=\sum_{k=0}^{24}\binom{24}{k}(-x^{3})^{24-k}(-3x^{-1})^{k}


    This should get you started. I have to go
    Last edited by TheEmptySet; April 3rd 2008 at 01:07 PM. Reason: La tex error
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  3. #3
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    I've tried using this, but I get the wrong answer in the end..
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  4. #4
    Oli
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    You need the term such that:
    (Ax^3)^a*(Bx^-1)^(24-a) is constant.

    This is only satisfied when a = 6

    Hence the constant term is:

    (-1)^6*(-3)^18*(24 choose 6)
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  5. #5
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    I'm back here is how you finish.

    <br />
([(-x)]^3+(-3x^{-1}))^{24}=\sum_{k=0}^{24}\binom{24}{k}(-x^{3})^{24-k}(-3x^{-1})^{k}<br />

    \sum_{k=0}^{24} \binom{24}{k}(-1)^{24-k}x^{72-3k}(-3)^{k}(x)^{-k}= \sum_{k=0}^{24} \binom{24}{k}(3)^{k}x^{72-4k}

    Sinc ewe want the constant term we want the exponent on the x to be zero.

    solving the equation

    72-4k=0 \iff k=18

    we can generate the constant term using k=18

    \binom{24}{18}(-x^3)^{24-18}(-3x^{-1})^{18}= \binom{24}{18}(-x^3)^6\left(\frac{-3}{x} \right)^{18}=\binom{24}{18}(3)^{18}=52125248137444
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  6. #6
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    thanks! got it now! =)
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