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

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(C) First,we resolve the expression into partial fractions: $\frac{x - 4}{x^2 - 5x + 6} = \frac{x - 4}{(x - 2)(x - 3)}$. Let $\frac{x - 4}{(x - 2)(x -

Vedclass · Accepted Answer

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

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(C) First,we resolve the expression into partial fractions: $\frac{x - 4}{x^2 - 5x + 6} = \frac{x - 4}{(x - 2)(x - 3)}$. Let $\frac{x - 4}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}$. $x - 4 = A(x - 3) + B(x - 2)$. For $x = 2$,$2 - 4 = A(2 - 3) \implies -2 = -A \implies A = 2$. For $x = 3$,$3 - 4 = B(3 - 2) \implies -1 = B \implies B = -1$. So,$\frac{x - 4}{x^2 - 5x + 6} = \frac{2}{x - 2} - \frac{1}{x - 3} = \frac{2}{-(2 - x)} - \frac{1}{-(3 - x)} = -\frac{2}{2(1 - x/2)} + \frac{1}{3(1 - x/3)} = -\frac{1}{1 - x/2} + \frac{1}{3(1 - x/3)}$. Using the expansion $(1 - y)^{-1} = \sum_{n=0}^{\infty} y^n$,we get: $-\sum_{n=0}^{\infty} (\frac{x}{2})^n + \frac{1}{3} \sum_{n=0}^{\infty} (\frac{x}{3})^n = -\sum_{n=0}^{\infty} \frac{x^n}{2^n} + \sum_{n=0}^{\infty} \frac{x^n}{3^{n+1}}$. The coefficient of $x^n$ is $-\frac{1}{2^n} + \frac{1}{3^{n+1}}$.

The coefficient of $x^n$ in the expansion of $\frac{x - 4}{x^2 - 5x + 6}$ in ascending powers of $x$ is:

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