Question

If the expansions in power of x of the function $\dfrac1{\left(1-\mathrm{ax}\right)\left(1-\mathrm{bx}\right)}\left(\mathrm a\neq\mathrm b\right)$ is a0 + a1x + a2x2 + ... then an is-
       $\mathrm A.\;\dfrac{\mathrm b^{\mathrm n}-\mathrm a^{\mathrm n}}{\mathrm b-\mathrm a}\\\mathrm B.\;\dfrac{\mathrm a^{\mathrm n}-\mathrm b^{\mathrm n}}{\mathrm b-\mathrm a}\\\mathrm C.\;\dfrac{\mathrm a^{\mathrm n+1}-\mathrm b^{\mathrm n+1}}{\mathrm b-\mathrm a}\\\mathrm D.\;\dfrac{\mathrm b^{\mathrm n+1}-\mathrm a^{\mathrm n+1}}{\mathrm b-\mathrm a}$

Answer 1

Hint: The expression for binomial expansion will be used in this question. The formula used is-
$\left(1-\mathrm x\right)^{-1}=1+\mathrm x+\mathrm x^2+..\;\mathrm{inifinite}\;\mathrm{terms}$

Complete step-by-step answer:
In the question, we have to find and predict the nth term which is the coefficient of xⁿ

$\dfrac1{\left(1-\mathrm{ax}\right)\left(1-\mathrm{bx}\right)}=\left(1-\mathrm{ax}\right)^{-1}\left(1-\mathrm{bx}\right)^{-1}\\=\left(1+\mathrm{ax}+\mathrm a^2\mathrm x^2+...+\mathrm a^{\mathrm n}\mathrm x^{\mathrm n}+...\right)\left(1+\mathrm{bx}+\mathrm b^2\mathrm x^2...+\mathrm b^{\mathrm n}\mathrm x^{\mathrm n}+...\right)\\\mathrm{We}\;\mathrm{have}\;\mathrm{to}\;\mathrm{find}\;\mathrm{the}\;\mathrm{coefficient}\;\mathrm{of}\;\mathrm x^{\mathrm n}\\=\mathrm b^{\mathrm n}+\mathrm b^{\mathrm n-1}\mathrm a+....+\mathrm{ba}^{\mathrm n-1}+\mathrm a^{\mathrm n}\\\\$
This term is a GP of n terms with common ratio $\dfrac{\mathrm a}{\mathrm b}$
We have to find the sum of this GP. The sum of a GP
The sum of a GP is given by the formula-
$\dfrac{\mathrm a\left(\mathrm r^{\mathrm n+1}-1\right)}{\mathrm r-1}\\\mathrm{Substituting}\;\mathrm{the}\;\mathrm{values}-\\\dfrac{\mathrm b^{\mathrm n}\left(\left({\displaystyle\dfrac{\mathrm a}{\mathrm b}}\right)^{\mathrm n+1}-1\right)}{{\displaystyle\dfrac{\mathrm a}{\mathrm b}}-1}\\=\dfrac{\mathrm b^{\mathrm n}}{\mathrm b^{\mathrm n+1}}\left(\dfrac{\mathrm a^{\mathrm n+1}-\mathrm b^{\mathrm n+1}}{\mathrm a-\mathrm b}\right)\mathrm b\\=\dfrac{\mathrm b^{\mathrm n+1}-\mathrm a^{\mathrm n+1}}{\mathrm b-\mathrm a}$
This is the answer.

Hence, the correct option is $\mathrm D.\;\dfrac{\mathrm b^{\mathrm n+1}-\mathrm a^{\mathrm n+1}}{\mathrm b-\mathrm a}$

Note: To solve this problem, one needs to have a knowledge of a GP and its sum. Also, to find the term an, we need to analyze the binomial expansion properly, and write the expression according to the pattern followed.