The coefficient of x^3 in the power series ex | vedclass

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(A) We use the generalized binomial expansion: $(1+z)^{-n} = 1 - nz + \frac{n(n+1)}{2!}z^2 - \frac{n(n+1)(n+2)}{3!}z^3 + \dots$For $(1+3x)^{-3}$,we ha

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-$27$

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(A) We use the generalized binomial expansion: $(1+z)^{-n} = 1 - nz + \frac{n(n+1)}{2!}z^2 - \frac{n(n+1)(n+2)}{3!}z^3 + \dots$For $(1+3x)^{-3}$,we have $n=3$ and $z=3x$:$(1+3x)^{-3} = 1 - 3(3x) + \frac{3(4)}{2}(3x)^2 - \frac{3(4)(5)}{6}(3x)^3 + \dots$$= 1 - 9x + 54x^2 - 270x^3 + \dots$Now,multiply by $(1+4x-3x^2)$:$(1+4x-3x^2)(1-9x+54x^2-270x^3 + \dots)$The coefficient of $x^3$ is obtained by:$1(-270) + 4(54) - 3(-9) = -270 + 216 + 27 = -27$.

The coefficient of $x^3$ in the power series expansion of $\frac{1+4x-3x^2}{(1+3x)^3}$ is

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