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Math Help - Triple term binomial expansion

  1. #1
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    Triple term binomial expansion

    In the expansion (1 + ax + bx^2)^5 the coefficients of x and x^2 are 10 and 25 respectively, where a and b are constants.

    Find the values of a and b.

    I have tried to rewrite the equation in the form ((1 + ax) + bx^2)^5 and prepare it for binomial expansion but I am getting something like
     \displaystyle 5 \choose 0 \displaystyle (1+ax)^5(bx^2)^0 + \displaystyle 5 \choose 1 \displaystyle (1+ax)^4(bx^2)^1 + ......

    Which we can use to solve for a and b but it is extremely long and tedious and I am sure there must be an easier way to get around it. Any help at all is appreciated.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Use:

    (u+v+w)^5=\displaystyle\sum_{\alpha +\beta +\gamma=5}\dfrac{5!}{\alpha!\beta!\gamma!}u^{\alph  a}v^{\beta}w^{\gamma}

    \alpha,\beta,\gamma non-negative integers.

    Fernando Revilla
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  3. #3
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    the trinomial theorem is not in my syllabus. is there another way around it?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by arccos View Post
    the trinomial theorem is not in my syllabus. is there another way around it?
    In that case, use the tedious method.

    Fernando Revilla
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  5. #5
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    \displaystyle (1+ax+bx^2)^5 = \sum_{0\le k \le 5}\binom{5}{k}(ax+bx^2)^k = 1+5(ax+bx^2)+10(ax+bx^2)^2+... .

    By inspection, the coefficient of x is 5a and that of x^2 is 10a^2+5b, thus we have:

    5a = 10 \Rightarrow a = 2 and 10a^2+b = 25 \Rightarrow 40+5b = 25 \Rightarrow b = -3.
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  6. #6
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    Quote Originally Posted by TheCoffeeMachine View Post
    \displaystyle (1+ax+bx^2)^5 = \sum_{0\le k \le 5}\binom{5}{k}(ax+bx^2)^k = 1+5(ax+bx^2)+10(ax+bx^2)^2+... .

    By inspection, the coefficient of x is 5a and that of x^2 is 10a^2+5b, thus we have:

    5a = 10 \Rightarrow a = 2 and 10a^2+b = 25 \Rightarrow 40+5b = 25 \Rightarrow b = -3.
    You should point out that the ellipsis ... contains only terms of order greater than 2.

    CB
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    You should point out that the ellipsis ... contains only terms of order greater than 2.
    I should have, indeed. Thanks.
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  8. #8
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    Quote Originally Posted by TheCoffeeMachine View Post
    \displaystyle (1+ax+bx^2)^5 = \sum_{0\le k \le 5}\binom{5}{k}(ax+bx^2)^k = 1+5(ax+bx^2)+10(ax+bx^2)^2+... .

    By inspection, the coefficient of x is 5a and that of x^2 is 10a^2+5b, thus we have:

    5a = 10 \Rightarrow a = 2 and 10a^2+b = 25 \Rightarrow 40+5b = 25 \Rightarrow b = -3.
    Oh yeah I could've grouped it another way... my bad. Haven't been doing expansions for a long time. Thanks!
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