If the expansion in powers of $x$ of the function $\frac{1}{(1-a x)(1-b x)}$ is $a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots$, then $a_n$ is

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Aicharya · Accepted Answer

$(1-a x)^{-1}(1-b x)^{-1}=\left(1+a x+a^2 x^2+\ldots \ldotsight)\left(1+b x+b^2 x^2+\ldots.ight)$ $	herefore$ coefficient of $x^n=b^n+a b^{n-1}+a^2 b^{n-2}+\ldots .+a^{n-1} b+a^n=\frac{b^{n+1}-a^{n+1}}{b-a}$ $	herefore a_n=\frac{b^{n+1}-a^{n+1}}{b-a}$

If the expansion in powers of $x$ of the function $\frac{1}{(1-a x)(1-b x)}$ is $a_0+a_1 x+a_2 x^2+a_3…

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