Question

If the expansion in powers of $x$ of the function $1/\left[\right(1-ax\left)\right(1-bx\left)\right]$ is $a_o+a_{1}x+a_2{x}^2+a_3{x}^3+...$, 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. $a_n=\frac{a^{n+1}-b^{n+1}}{a-b}$

Answer 1

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

$(1-ax{)}^{-1}(1-bx{)}^{-1}$

$=[1+ax+a^2{x}^2+a^3{x}^3...\infty ][1+bx+b^2{x}^2+b^3{x}^3...\infty ]$

$=[1+(a+b)x+(a^2+ab+b^2)x^2+...\infty ]$

Comparing coefficients, we get

$a_0=1$

$a_1=(a+b)=\frac{a^2-b^2}{a-b}$

$a_2=(a^2+ab+b^2)=\frac{a^3-b^3}{a-b}$
:
:
$a_n=\frac{a^{n+1}-b^{n+1}}{a-b}$