Question

If the expansion of $\frac{1}{(1-ax)(1-bx)}=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n+\cdots$, then $a_n$ is (where $a\ne b,|ax|,|bx|<1$)

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

$\frac{1}{(1-ax)(1-bx)}=a_0+a_{1}x+a_2{x}^2+.....+a_n{x}^n+....$
but $(1-ax{)}^{-1}(1-bx{)}^{-1}=(1+ax+a^2{x}^2+\cdots )(1+bx+b^2{x}^2+\cdots )$
$\Rightarrow$ comparing the coefficient of $x^n$ in above equations
$b^n+a{b}^{n-1}+a^2{b}^{n-2}+....+a^{n-1}b+a^n=a_n$
$\therefore a_n=\frac{b^{n+1}-a^{n+1}}{b-a}$