The coefficient of x^n in the expansion of fr | vedclass

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Question

(C) Given expression: $\frac{1}{x^2-5x+6} = \frac{1}{(x-3)(x-2)}$.Using partial fractions: $\frac{1}{(x-3)(x-2)} = \frac{1}{x-3} - \frac{1}{x-2} = \fr

Vedclass · Accepted Answer

$\frac{1}{2^{n+1}}-\frac{1}{3^{n+1}}$

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(C) Given expression: $\frac{1}{x^2-5x+6} = \frac{1}{(x-3)(x-2)}$.Using partial fractions: $\frac{1}{(x-3)(x-2)} = \frac{1}{x-3} - \frac{1}{x-2} = \frac{1}{2-x} - \frac{1}{3-x}$.Rewrite the expression as: $\frac{1}{2(1-\frac{x}{2})} - \frac{1}{3(1-\frac{x}{3})}$.Using the binomial expansion $(1-y)^{-1} = \sum_{n=0}^{\infty} y^n$ for $|y| $= \frac{1}{2} \sum_{n=0}^{\infty} (\frac{x}{2})^n - \frac{1}{3} \sum_{n=0}^{\infty} (\frac{x}{3})^n$.The coefficient of $x^n$ is $\frac{1}{2} \cdot (\frac{1}{2})^n - \frac{1}{3} \cdot (\frac{1}{3})^n = \frac{1}{2^{n+1}} - \frac{1}{3^{n+1}}$.

The coefficient of $x^n$ in the expansion of $\frac{1}{x^2-5x+6}$ for $|x| < 1$ is

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