Question

If $\frac{1}{(1-2x)(1+3x)}$ is to be expanded as a power series of $x$, then

• A. $|x|<1/2$
• B. $|x|<1/6$
• C. $\frac{1}{3}$
• D. $|x|<1/3$

Answer 1

Answer: (B)

$|x|<1/6$

$\begin{array}{c}\frac{1}{\left(1-2x\right)\left(1+3x\right)}\\ \Rightarrow {\left(1-2x\right)}^{-1}{\left(1+3x\right)}^{-1}=\frac{1}{2}\cdot \frac{1}{3}{\left(\frac{1}{2}-x\right)}^{-1}{\left(\frac{1}{3}+x\right)}^{-1}\\ \Rightarrow byusingbinomialofrationalindex\\ \Rightarrow {\left(1+x\right)}^n\\ \left|x\right|<1\\ Hence,\\ \Rightarrow \left|x\right|<\frac{1}{2}\\ \Rightarrow \left|x\right|<\frac{1}{3}\\ \therefore \left|x\right|<\frac{1}{6}\end{array}$