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

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Question

(A) Using partial fractions,we write: $\frac{5x + 6}{(2 + x)(1 - x)} = \frac{A}{2 + x} + \frac{B}{1 - x}$.Solving for $A$ and $B$: $5x + 6 = A(1 - x)

Vedclass · Accepted Answer

$\frac{-2}{3} \frac{(-1)^n}{2^n} + \frac{11}{3}$

Vedclass · Answer

(A) Using partial fractions,we write: $\frac{5x + 6}{(2 + x)(1 - x)} = \frac{A}{2 + x} + \frac{B}{1 - x}$.Solving for $A$ and $B$: $5x + 6 = A(1 - x) + B(2 + x)$.For $x = 1$: $11 = 3B \implies B = \frac{11}{3}$.For $x = -2$: $-4 = 3A \implies A = -\frac{4}{3}$.Thus,$\frac{5x + 6}{(2 + x)(1 - x)} = \frac{-4/3}{2 + x} + \frac{11/3}{1 - x} = \frac{-2/3}{1 + x/2} + \frac{11/3}{1 - x}$.Expanding as a power series: $\frac{-2}{3}(1 + x/2)^{-1} + \frac{11}{3}(1 - x)^{-1}$.$= \frac{-2}{3} \sum_{n=0}^{\infty} (-\frac{x}{2})^n + \frac{11}{3} \sum_{n=0}^{\infty} x^n$.$= \sum_{n=0}^{\infty} [\frac{-2}{3} \frac{(-1)^n}{2^n} + \frac{11}{3}] x^n$.The coefficient of $x^n$ is $\frac{-2}{3} \frac{(-1)^n}{2^n} + \frac{11}{3}$.

The coefficient of $x^n$ in the expansion of $\frac{5x + 6}{(2 + x)(1 - x)}$ in ascending powers of $x$ is:

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