If the expansion in power of $x$ of the function $\frac{1}{(1 - ax)(1 - bx)}\ is\ a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ...,\ then\ a_{n}\ is$

Question

Dr. Abubacker · Accepted Answer

(c)
$(1 - a\ x)^{- 1}(1 - bx)^{- 1}$
$= (a^{0} + a\ x + a^{2}x^{2} +$…$(b^{0} + bx + b^{2}x^{2} +$….)
Hence, $a_{n} =$coefficient of $x^{n}$ in $(1 - ax)^{- 1}(1 - bx)^{- 1}$
$a^{0}b^{n} + {ab}^{n - 1} +$…..+$a^{n}b^{0}$
$= a^{0}b^{n}\left( 1 + \frac{a}{b} + \left( \frac{a}{b} \right)^{2} + .\ldots + \left( \frac{a}{b} \right)^{n} \right)$
$= a^{0}b^{n}\left( \frac{\left( \frac{a}{b} \right)^{n + 1} - 1}{\frac{a}{b} - 1} \right)$
$= \frac{a^{n + 1} - b^{n + 1}}{a - b} = \frac{b^{n + 1} - a^{n + 1}}{b - a}$

Q. If the expansion in power of $x$ of the function $\frac{1}{(1 - ax)(1 - bx)}\ is\ a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ...,\ then\ a_{n}\ is$

Explanation

Reviewed by Dr. Abubacker

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