Question

If the expansion of powers of x of the function $\frac{1}{(1-ax)(1-bx)}$ is $a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots ,$ then $a_n$ is?

• A. $\frac{b^n-a^n}{b-a}$
• B. $\frac{a^n-b^n}{b-a}$
• 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 D

$\frac{b^{n+1}-a^{n+1}}{b-a}$

$(1-ax{)}^{-1}(1-bx{)}^{-1}=(1+ax+a^2{x}^2+\cdots )(1+bx+b^2{x}^2+\cdots )$ therefore, the coefficient of $x^n$ is
$b^n+a{b}^{n-1}+a^2{b}^{n-2}+\cdots a^{n-1}b+a^n=\frac{b^{n+1}-a^{n+1}}{b-a}\therefore a_n=\frac{b^{n+1}-a^{n+1}}{b-a}$