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Mathematics Question on Sequence and series

If the expansion in powers of xx of the function 1(1−ax)(1−bx)\frac{1}{\left(1-ax\right)\left(1-bx\right)} is a0+a1x+a2x2+a3x3+.....,a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+ ..... , then ana_n is :

A

an−bnb−a\frac{a^{n}-b^{n}}{b-a}

B

an+1−bn+1b−a\frac{a^{n+1}-b^{n+1}}{b-a}

C

bn+1−an+1b−a\frac{b^{n+1}-a^{n+1}}{b-a}

D

bn−anb−a\frac{b^{n}-a^{n}}{b-a}

Answer

bn+1−an+1b−a\frac{b^{n+1}-a^{n+1}}{b-a}

Explanation

Solution

∵(1−ax)−1(1−bx)−1\because\left(1-ax\right)^{-1}\left(1-bx\right)^{-1} =(1+ax+a2x2+...)(1+bx+b2x2+...)=\left(1+ax+a^{2}x^{2}+...\right)\left(1+bx+b^{2}x^{2}+...\right) Hence an=a_n = coefficient of xnx^n in (1−ax)−1(1−bx)−1\left(1-ax\right)^{-1}\left(1-bx\right)^{-1} =a0bn+abn−1+...+anb0=a^{0}b^{n}+ab^{n-1}+...+a^{n}b^{0} =a0bn((ab)n+1−1ab−1)R=a^{0}b^{n}\left(\frac{\left(\frac{a}{b}\right)^{n+1}-1}{\frac{a}{b}-1}\right)R =bn(an+1−bn+1)a−b⋅bbn+1=an+1−bn+1a−b=\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}