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6 November, 03:53

What is the numerical coefficient of the a^8*b^2 term in the expansion of ((1/3) a^2 - 3b) ^6?

Enter your answer in simplest fractional form.

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  1. 6 November, 04:13
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    In the binomial development, the main problem is calculation of binomial coefficients.

    If we want to get term a∧8*b∧2 we see that this is the third member in binomial development (n 2) a∧n-2*b∧2

    The given binomial is ((1/3) a∧2 - 3b) ∧6, the first element is (1/3) a∧2, the second element is (-3b) and n=6 when we replace this in the formula we get

    (6 2) * ((1/3) a∧2) ∧ (6-2) * (-3b) 2 = (6*5) / 2 * ((1/3) a∧2) ∧4 * 9b∧2 = 15 * (1/81) * 9 * (a∧8b∧2) =

    = 15*9 * a∧8b∧2 = 135*a∧8b∧2

    We finally get numerical coefficient 135

    Good luck!
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