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Math Help - TOUGH binomial theorem/problem

  1. #1
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    TOUGH binomial theorem/problem

    when (1 + ax + bx^2 + cx^3)^10 is expanded in ascending powers of x, the coefficients of x, x^2 , x^3 are 20, 200, 1000 respectively. Find the values of a, b , c and the coefficient of x^4.

    Racked my brain and the internet looking for answers. Its for college algebra :S! any and all help appreciated thanks.
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  2. #2
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    Are you SURE this is a binomial therorem problem? If so, I suppose you could find a factorization that would mike your life easier.

    For my money, it seemed quite a bit easier to create the McLauren expansion and solve sequentially for a, b, and c. It's only a couple of derivatives.
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  3. #3
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    yeah supposedly we have to use binomial theorem. My best guess is that u just find eliminate the a b c and find the coefficients for x, xsqaured and xcubed. and i tried that but im not sure if what i got was right since i only did 10C2 = 45 and since 20 = the final coefficient you just 20/45=a. however since u have so many terms in the ^10 i doubt thats what it is. since it would work for any (a+b)^10 but not a ((1+ax)+(bx^2+cx^3))
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  4. #4
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    To expand \left( {1 + ax + bx^2  + cx^3  + dx^4 } \right)^{10} we would have ten factors. The only way to have a term containing x is to use the ax in exactly one of the factors and 1’s all the other factors. That will give us 10ax, so a=2. To have a term containing x^2 there are two ways to construct it. First, we could take ax from exactly two of the factors and 1’s else. That can be done in {{10} \choose {2}}=45 ways. Or we can take bx^2 from exactly one of the factors and 1’s else. Again that can be done in 10 ways. So we end up with 45a^2x^2 + 10bx^2. So we know 180+10b=200, solve for b.

    Now you must continue with the x^3 case. How can it be made up?
    Then proceed to the x^4 case. How can it be made up?
    Last edited by Plato; September 8th 2008 at 11:25 AM.
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  5. #5
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    I'm not quite sure how using a five-element base constitutes using the Binaomial Theorem. Anyway, since you're going only up to x^4, you won't be using the original x^3 term very much. In other words, it shouldn't be hard to track down the few terms you'll need.
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