Question

If the expression in powers of $x$ of the function $\frac{1}{\left[(1-ax)(1-bx)\right]}$ is $a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\dots ..$then the coefficient of$x^n$ in the expansion is

• A. $\frac{\left[a^n–b^n\right]}{b–a}$
• B. $\frac{\left[a^{n+1}–b^{n+1}\right]}{a-b}$
• C. $\frac{\left[b^{n+1}–a^{n+1}\right]}{a-b}$
• D. $\frac{\left[b^n–a^n\right]}{b–a}$

Answer 1

The correct option is B

$\frac{\left[a^{n+1}–b^{n+1}\right]}{a-b}$

Explanation for the correct option:

Given function is $\frac{1}{\left[(1-ax)(1-bx)\right]}={(1-ax)}^{-1}{(1-bx)}^{-1}$

We know that the binomial expansion for ${\left(1-x\right)}^{-1}=1+x+x^2+x^3+...\infty$

$\begin{array}{cccl}\Rightarrow & {(1-ax)}^{-1}{(1-bx)}^{-1}& =& \left(1+ax+{\left(ax\right)}^2+{\left(ax\right)}^3+....\infty \right)\left(1+bx+{\left(bx\right)}^2+{\left(bx\right)}^3+....\infty \right)\end{array}$

Therefore, coefficients of $x^n$ in the above expansion can be

$\begin{array}{cl}& \left(a^0{b}^n+a^1{b}^{n-1}+a^2{b}^{n-2}+a^3{b}^{n-3}+.....+a^{n-1}{b}^1+a^n{b}^0\right)\\ =& b^n\left(1+\frac{a}{b}+{\left(\frac{a}{b}\right)}^2+{\left(\frac{a}{b}\right)}^3+.....+{\left(\frac{a}{b}\right)}^{n-1}+{\left(\frac{a}{b}\right)}^n\right)\end{array}$

The above expression is in the form of $n$terms of a Geometric progression with first term $p$as $1$ and common ratio $r$ as $\frac{a}{b}$

The general formula for sum of $n$terms of a Geometric progression is $\frac{p\left(r^n-1\right)}{r-1}$

$\begin{array}{cccl}\therefore & b^n\left(1+\frac{a}{b}+{\left(\frac{a}{b}\right)}^2+{\left(\frac{a}{b}\right)}^3+.....+{\left(\frac{a}{b}\right)}^{n-1}+{\left(\frac{a}{b}\right)}^n\right)& =& b^n\left(\frac{{\left({\displaystyle \frac{a}{b}}\right)}^{n+1}-1}{{\displaystyle \frac{a}{b}}-1}\right)\\ & & =& b^n\left(\frac{{\displaystyle \frac{a^{n+1}-b^{n+1}}{b^{n+1}}}}{{\displaystyle \frac{a-b}{b}}}\right)\\ & & =& b^n\left(\frac{a^{n+1}-b^{n+1}}{b^n\left(a-b\right)}\right)\\ \Rightarrow & b^n\left(1+\frac{a}{b}+{\left(\frac{a}{b}\right)}^2+{\left(\frac{a}{b}\right)}^3+.....+{\left(\frac{a}{b}\right)}^{n-1}+{\left(\frac{a}{b}\right)}^n\right)& =& \frac{a^{n+1}-b^{n+1}}{\left(a-b\right)}\end{array}$

Hence, the correct answer is option(B) i.e. $\frac{\left[a^{n+1}–b^{n+1}\right]}{a-b}$.