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 $\frac{15}{8}$

$\frac{2}{(1-x)(2-x)}=[\frac{1}{1-x}-\frac{1}{2-x}]\times 2=2\times [{(1-x)}^{-1}-{(1-\frac{x}{2})}^{-1}\times \frac{1}{2}]$

Coefficient of $x^r$ in $(1+x{)}^n$ is $\frac{n.(n-1)(n-2)...(n-r+1)x^r}{r!}$

Hence, Coefficient of $x^3=$

$2\times (\frac{(-1).(-2)(-3){(-1)}^3}{3!}-\frac{(-1).(-2)(-3){\left(\frac{-1}{2}\right)}^3}{3!\times 2})$

$=2\times (1-\frac{1}{8\times 2})=\frac{15}{8}$