The coefficient of x^{3} in the infinite seri | vedclass

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(B) We can use partial fractions to decompose the expression: $\frac{2}{(1-x)(2-x)} = \frac{A}{1-x} + \frac{B}{2-x}$.Solving for $A$ and $B$: $2 = A(2

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$\frac{15}{8}$

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(B) We can use partial fractions to decompose the expression: $\frac{2}{(1-x)(2-x)} = \frac{A}{1-x} + \frac{B}{2-x}$.Solving for $A$ and $B$: $2 = A(2-x) + B(1-x)$.For $x=1$, $2 = A(1) \implies A=2$.For $x=2$, $2 = B(-1) \implies B=-2$.So, $\frac{2}{(1-x)(2-x)} = \frac{2}{1-x} - \frac{2}{2-x} = 2(1-x)^{-1} - (1-\frac{x}{2})^{-1}$.Expanding using the binomial series $(1-z)^{-1} = 1 + z + z^2 + z^3 + \dots$:$2(1 + x + x^2 + x^3 + \dots) - (1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \dots)$.The coefficient of $x^3$ is $2(1) - \frac{1}{8} = 2 - \frac{1}{8} = \frac{16-1}{8} = \frac{15}{8}$.

The coefficient of $x^{3}$ in the infinite series expansion of $\frac{2}{(1-x)(2-x)},$ for $|x| < 1,$ is

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