Question

If the expansion in powers of $x$ of the function $\frac{1}{\left(1-ax\right)\left(1-bx\right)}$ is
$a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+ ..... ,$ then $a_n$ is :

• A. $\frac{a^{n}-b^{n}}{b-a}$
• B. $\frac{a^{n+1}-b^{n+1}}{b-a}$
• C. $\frac{b^{n+1}-a^{n+1}}{b-a}$
• D. $\frac{b^{n}-a^{n}}{b-a}$

Answer 1

Answer: (C)

$\because\left(1-ax\right)^{-1}\left(1-bx\right)^{-1}$
$=\left(1+ax+a^{2}x^{2}+...\right)\left(1+bx+b^{2}x^{2}+...\right)$
Hence $a_n =$ coefficient of $x^n$ in $\left(1-ax\right)^{-1}\left(1-bx\right)^{-1}$
$=a^{0}b^{n}+ab^{n-1}+...+a^{n}b^{0}$
$=a^{0}b^{n}\left(\frac{\left(\frac{a}{b}\right)^{n+1}-1}{\frac{a}{b}-1}\right)R$
$=\frac{b^{n}\left(a^{n+1}-b^{n+1}\right)}{a-b}\cdot\frac{b}{b^{n+1}}=\frac{a^{n+1}-b^{n+1}}{a-b}$