Question

If $\frac{1}{(1-ax)(1-bx)}=a_0+a_{1}x+a_2{x}^2+...then{a}_n$ equals

• 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 $\frac{b^{n+1}-a^{n+1}}{b-a}$

$\frac{1}{(1-ax)(1-bx)}=a_0+a_{1}x+a_2{x}^2+...$
$\Rightarrow \frac{1}{x(a-b)}[\frac{1}{1-ax}-\frac{1}{1-bx}]=a_0+a_{1}x+a_2{x}^2+...$
$\Rightarrow \frac{1}{x(a-b)}[1+ax+a^2{x}^2+..+a^{n-1}{x}^{n-1}+..\infty -(1+bx+b^2{x}^2+..+b^{n-1}{x}^{n-1}+..\infty )]=a_0+a_{1}x+a_2{x}^2+...$
$\Rightarrow \frac{(a-b)+(a^2-b^2)x+(a^3-b^3)x^2+....+(a^{n+1}-b^{n+1})x^n+...\infty }{a-b}=a_0+a_{1}x+a_2{x}^2+...$
Therefore, $a_n=\frac{b^{n+1}-a^{n+1}}{b-a}$