If the expansion in powers of x of the functi | vedclass

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Question

$\frac{1}{(1-a x)(1-b x)}$

$=(1-a x)^{-1}(1-b x)^{-1}$

$=\left[1+(-1)(-a x)+\frac{(-1)(-2)}{1.2}(-a x)^{2}+\ldots\right]$$\left[1+(-1)(-b x)+\fr

Vedclass · Accepted Answer

$\;\frac{{{b^{n + 1}} - {a^{n + 1}}}}{{b - a}}$

Vedclass · Answer

$\frac{1}{(1-a x)(1-b x)}$

$=(1-a x)^{-1}(1-b x)^{-1}$

$=\left[1+(-1)(-a x)+\frac{(-1)(-2)}{1.2}(-a x)^{2}+\ldots\right]$$\left[1+(-1)(-b x)+\frac{(-1)(-2)}{1.2}(-b x)^{2}+\ldots\right]$

$=\left[1+a x+a^{2} x^{2}+\ldots+a^{n-1} x^{n-1}+a^{n} x^{n}+\ldots .\right]$$\left[1+b x+b^{2} x^{2}+\ldots+b^{n-1} x^{n-1}+b^{n} x^{n}+\ldots\right]$

Coefficient of $x^{n}=$

$a^{n}+a^{n-1} b+a^{n-2} b^{2}+\ldots .+b^{n}$

$=a^{n}\left[1+\frac{b}{a}+\frac{b^{2}}{a^{2}}+\ldots \ldots+\frac{b^{n}}{a^{n}}\right]$

$=a^{n}\left[\frac{\left(\frac{b}{a}\right)^{n+1}-1}{\frac{b}{a}-1}\right]$

$=a^{n}\left[\frac{b^{n+1}-a^{n+1}}{a^{n+1}\left(\frac{b-a}{a}\right)}\right]$

$=\frac{b^{n+1}-a^{n+1}}{b-a}$

If the expansion in powers of $x$ of the function $\frac{1}{{\left( {1 - ax} \right)\left( {1 - bx} \right)}}$ is ${a_0} + {a_1}x + {a_2}{x^2} + \;{a_3}{x^3} + \; \ldots......$ then ${a_n}$ is

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