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

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(A) We use partial fractions to decompose the expression: $\frac{1}{(1 - x)(3 - x)} = \frac{A}{1 - x} + \frac{B}{3 - x}$ Solving for $A$ and $B$,we ge

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

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(A) We use partial fractions to decompose the expression: $\frac{1}{(1 - x)(3 - x)} = \frac{A}{1 - x} + \frac{B}{3 - x}$ Solving for $A$ and $B$,we get $1 = A(3 - x) + B(1 - x)$. For $x = 1$,$1 = 2A \implies A = 1/2$. For $x = 3$,$1 = -2B \implies B = -1/2$. Thus,$\frac{1}{(1 - x)(3 - x)} = \frac{1}{2} \left( \frac{1}{1 - x} - \frac{1}{3 - x} \right) = \frac{1}{2} \left[ (1 - x)^{-1} - \frac{1}{3} (1 - x/3)^{-1} \right]$. Using the binomial expansion $(1 - z)^{-1} = \sum_{n=0}^{\infty} z^n$: $\frac{1}{2} \left[ \sum_{n=0}^{\infty} x^n - \frac{1}{3} \sum_{n=0}^{\infty} (x/3)^n \right] = \frac{1}{2} \sum_{n=0}^{\infty} \left( 1 - \frac{1}{3^{n+1}} \right) x^n$. The coefficient of $x^n$ is $\frac{1}{2} \left( 1 - \frac{1}{3^{n+1}} \right) = \frac{3^{n+1} - 1}{2 \cdot 3^{n+1}}$.

The coefficient of ${x^n}$ in the expansion of $\frac{1}{(1 - x)(3 - x)}$ is

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