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Thread: How to use binomial theorem to work out the coefficient of s^27

  1. #1
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    How to use binomial theorem to work out the coefficient of s^27

    Hi, I have the following problem that is solved, but I have no clue what formula the teacher uses to solve it... Thanks a lot for your help!

    I have:

    $\displaystyle (((1/6)*s)^{10})*((1-s^6)/(1-s))^{10}$

    Which is equal to

    $\displaystyle ((1/6)^{10})*(1-10s^6+...)*(1+10s+...)$

    And the coefficient of $\displaystyle s^{27}$ (and the main thing I don't understand how the teacher gets it) is

    $\displaystyle [(1/6)^{10}]*[(10C2)*(14C5)-(10C1)*(20C11)+(26C17)]$
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  2. #2
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    Re: How to use binomial theorem to work out the coefficient of s^27

    Quote Originally Posted by buenogilabert View Post
    $\displaystyle (((1/6)*s)^{10})*((1-s^6)/(1-s))^{10}$

    Which is equal to

    $\displaystyle ((1/6)^{10})*(1-10s^6+...)*(1+10s+...)$
    The last line should be multiplied by $\displaystyle s^{10}$.

    By the simplest form of the binomial theorem we have

    $\displaystyle (1-s^6)^{10}=1-\binom{10}{1}s^6+\binom{10}{2}s^{12}+\dots$ (1)

    Also, by the last formula in this Wiki section, we have

    (2)

    We have $\displaystyle s^{10}$ that comes from the first factor $\displaystyle ((1/6)*s)^{10}$, so the rest of the expression must contribute $\displaystyle s^{17}$ towards $\displaystyle s^{27}$. We can get $\displaystyle s^{12}$ from (1) and $\displaystyle s^{5}$ from (2); $\displaystyle s^{6}$ from (1) and $\displaystyle s^{11}$ from (2); and $\displaystyle s^{0}$ from (1) and $\displaystyle s^{17}$ from (2). The product of the corresponding coefficients make the three terms in the expression

    $\displaystyle \binom{10}{2}\binom{14}{5}-\binom{10}{1}\binom{20}{11}+\binom{26}{17}$
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    Re: How to use binomial theorem to work out the coefficient of s^27

    I spent 80 minutes with my tutor and he didn't know how to do it... Thanks a lot!
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    Re: How to use binomial theorem to work out the coefficient of s^27

    Quote Originally Posted by buenogilabert View Post
    $\displaystyle (((1/6)*s)^{10})*((1-s^6)/(1-s))^{10}$
    And the coefficient of $\displaystyle s^{27}$
    Here is what I would do.
    Because $\displaystyle \left( {\frac{s}{6}} \right)^{10} = \frac{{s^{10} }}{{6^{10} }}$

    we are looking for the coefficient of $\displaystyle s^{17}$ in
    $\displaystyle \left( {\frac{{1 - s^6 }}{{1 - s}}} \right)^{10} = \left( {s^5 + s^4 + s^3 + s^2 + s + 1} \right)^{10} $.

    That is $\displaystyle 1535040x^{17}$.

    I found that using a CAS. I don't know how much you know about generating polynomials. So this may not help you at all.
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