(D) We have,$\frac{(1-3 x)^2}{(1-2 x)} = (1 - 6x + 9x^2)(1 - 2x)^{-1}$.Using the binomial expansion $(1 - y)^{-1} = 1 + y + y^2 + y^3 + y^4 + \dots$,w

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(D) We have,$\frac{(1-3 x)^2}{(1-2 x)} = (1 - 6x + 9x^2)(1 - 2x)^{-1}$.Using the binomial expansion $(1 - y)^{-1} = 1 + y + y^2 + y^3 + y^4 + \dots$,we get $(1 - 2x)^{-1} = 1 + 2x + 4x^2 + 8x^3 + 16x^4 + \dots$.So,the expression is $(1 - 6x + 9x^2)(1 + 2x + 4x^2 + 8x^3 + 16x^4 + \dots)$.To find the coefficient of $x^4$,we multiply terms:$1 \times (16x^4) = 16x^4$$-6x \times (8x^3) = -48x^4$$9x^2 \times (4x^2) = 36x^4$Summing these coefficients: $16 - 48 + 36 = 4$.

Class 11 Mathematics Binomial Theorem General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

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

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