The coefficient of x^n in the expression frac | 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)

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

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(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,the expression is $\frac{-4/3}{2 + x} + \frac{11/3}{1 - x} = \frac{-2/3}{1 + x/2} + \frac{11/3}{1 - x}$.Expanding using the binomial series $(1 + z)^{-1} = \sum_{n=0}^{\infty} (-1)^n z^n$ and $(1 - z)^{-1} = \sum_{n=0}^{\infty} z^n$:$= -\frac{2}{3} \sum_{n=0}^{\infty} (-1)^n \left(\frac{x}{2}\right)^n + \frac{11}{3} \sum_{n=0}^{\infty} x^n$.The coefficient of $x^n$ is $-\frac{2}{3} \cdot \frac{(-1)^n}{2^n} + \frac{11}{3}$.

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

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