Question

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

• A. $\frac{–1}{16}$
• B. $\frac{15}{8}$
• C. $\frac{–1}{8}$
• D. $\frac{15}{16}$

Answer 1

The correct option is B

$\frac{15}{8}$

Explanation for the correct option:

Finding coefficient of $x^3$:

Using infinite expansion series where

Given, $\frac{2}{(1-x)(2-x)}$, for $\left|x\right|<1$

Using series which is One of the most fundamental subjects in Arithmetic is sequence and series.
A sequence is an itemized collection of elements that allows for any type of repetition, whereas a series is the total of all elements.
One of the most common instances of sequence and series is an arithmetic progression.

Rearranging the series gives

$$\begin{gathered} 2{(1-x)}^{-1}{(2-x)}^{-1}=\frac{2}{2}{\left(1-x\right)}^{-1}{\left(1-\frac{x}{2}\right)}^{-1} \\ ={\left(1-x\right)}^{-1}{\left(1-\frac{x}{2}\right)}^{-1} \end{gathered}$$

The standard expansion of ${(1-x)}^{-n}$ is given by

${(1-x)}^{-1}=1+x+x^2+x^{3+...}$

$\Rightarrow {\left(1-x\right)}^{-1}{\left(1-\frac{x}{2}\right)}^{-1}=\left[1+x+x^2+........\right]\left[1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{8}+.....\right]$

Coefficient of $x^3$ in expression will be

$$\begin{gathered} =1·\frac{1}{8}+1·\frac{1}{4}+1·\frac{1}{2}+1·1 \\ =\frac{1+2+4+8}{8} \\ =\frac{15}{8} \end{gathered}$$

Hence, option (B) is the correct answer.