Question

If the expansion in powers of x of the function $\frac{1}{(1-ax)(1-bx)}$ is $a_o+a_{1}x+a_2{x}^2+$..., then $a_o$ is

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

Answer 1

The correct option is D $\frac{b^{n+1}-a^{n+1}}{b-a}$

$\frac{1}{(1-ax)(1-bx)}=(1-ax{)}^{-1}(1-bx{)}^{-1}$

$=(1+ax+a^2{x}^2....a^n{x}^n..)(1+bx+b^2{x}^2...b^n{x}^n...)$

(using Bionomial expansion)

$a_n=b^n+b^{n-1}a+b^{n-2}{a}^2.....a^n$

A GP with $a_0=b^n$ and $r=\frac{a}{b}$

$a_n=b^n\frac{(1-{\left(\frac{a}{b}\right)}^{n+1})}{1-\frac{a}{b}}$

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