Question

If the expansion in powers of x of the function $1/\left[\right(1-ax\left)\right(1-bx\left)\right]$ is $a_0+a_{1}x+a_2{x}^2+a_3{x}^2+$..., then $a_n$ is?

• A. $\frac{b^n-a^n}{b-a}$
• B. $\left(a^n+b^n\right)x^n$
• C. $\frac{a^{n+1}-b^{n+1}}{b-a}$
• D. $\frac{b^{n+1}-a^{n+1}}{b-a}$

Answer 1

The correct option is B $\left(a^n+b^n\right)x^n$

$\begin{array}{c}\frac{1}{\left(1-ax\right)\left(1-bx\right)}\\ \Rightarrow {\left(1-ax\right)}^{-1}{\left(1-bx\right)}^{-1}\\ Bybinomialexpansion\\ \Rightarrow \left[1+ax+\frac{1\times 2a^2{x}^2}{2!}+\frac{1\times 2\times 3a^3{x}^3}{3!}+............\frac{n!a^n{x}^n}{n!}\right]\times \\ \Rightarrow \left[1+bx+\frac{1\times 2b^2{x}^2}{2!}+\frac{1\times 2\times 3b^3{x}^3}{3!}+............\frac{n!b^n{x}^n}{n!}\right]\\ \Rightarrow \left(1+ax+a^2{x}^2+a^3{x}^3+........a^n{x}^n\right)\left(1+bx+b^2{x}^2+b^3{x}^3+........b^n{x}^n\right)\\ \Rightarrow 1+\left(a+b\right)x+\left(a^2+b^2\right)x^2+\left(a^3+b^3\right)x^3+...........\left(a^n+b^n\right)x^n\\ \therefore a_n=\left(a^n+b^n\right)x^{n}Ans.\end{array}$