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Question: If the expansions in power of x of the function \(\dfrac1{\left(1-\mathrm{ax}\right)\left(1-\mathrm{...

If the expansions in power of x of the function 1(1−ax)(1−bx)(a≠b)\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-
A.  bn−anb−aB.  an−bnb−aC.  an+1−bn+1b−aD.  bn+1−an+1b−a\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}

Explanation

Solution

Hint: The expression for binomial expansion will be used in this question. The formula used is-
(1−x)−1=1+x+x2+..  inifinite  terms\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 ab\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-
a(rn+1−1)r−1Substituting  the  values−bn((ab)n+1−1)ab−1bnbn+1ˉ(an+1−bn+1a−b)bbn+1−an+1b−aˉ\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 D.  bn+1−an+1b−a\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.