The coefficient of {x^n} in the expansion of | vedclass

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(B) We have,$(1 - 9x + 20{x^2})^{-1} = [(1 - 5x)(1 - 4x)]^{-1}$.Using partial fractions,we write:$\frac{1}{(1 - 5x)(1 - 4x)} = \frac{A}{1 - 5x} + \fra

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${5^{n + 1}} - {4^{n + 1}}$

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(B) We have,$(1 - 9x + 20{x^2})^{-1} = [(1 - 5x)(1 - 4x)]^{-1}$.Using partial fractions,we write:$\frac{1}{(1 - 5x)(1 - 4x)} = \frac{A}{1 - 5x} + \frac{B}{1 - 4x}$.Solving for $A$ and $B$,we get $A = 5$ and $B = -4$.Thus,the expression becomes:$5(1 - 5x)^{-1} - 4(1 - 4x)^{-1}$.Expanding using the binomial series $(1 - z)^{-1} = \sum_{k=0}^{\infty} z^k$:$= 5 \sum_{k=0}^{\infty} (5x)^k - 4 \sum_{k=0}^{\infty} (4x)^k$.The coefficient of ${x^n}$ is:$5(5^n) - 4(4^n) = 5^{n+1} - 4^{n+1}$.

The coefficient of ${x^n}$ in the expansion of ${(1 - 9x + 20{x^2})^{-1}}$ is

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