If $\frac{(1-p x)^{-1}}{(1-q x)}=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots$, then $a_n=$

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Aicharya · Accepted Answer

Since, $\begin{aligned} & \frac{(1-p x)^{-1}}{(1-q x)}=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots \ & \because(1-p x)^{-1}=1+p x+p^2 x^2+p^3 x^3+\ldots+p^n x^n+\ldots \end{aligned}$ and $(1-q x)^{-1}=1+q x+q^2 x^2+q^3 x^3+\ldots+q^n x^n+\ldots$ Now, coefficient of $x^n$ in the expansion of $\begin{aligned} & (1-p x)^{-1}(1-q x)^{-1} \ & =p^n+p^{n-1} q+p^{n-2} q^2+p^{n-3} q^3+\ldots+q^n \end{aligned}$ $\begin{aligned} & a_n=\frac{p^n\left(1-\left(\frac{q}{p}\right)^{n+1}\right)}{1-\frac{q}{p}}=\frac{p^n\left(p^{n+1}-q^{n+1}\right) p}{(p-q) p^{n+1}} \ & =\frac{p^{n+1}-q^{n+1}}{p-q} \ & \text{So, } a_n=\frac{p^{n+1}-q^{n+1}}{p-q} \ \end{aligned}$ Hence, option (2) is correct.

If \(\frac{(1-p x)^{-1}}{(1-q x)}=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots\), then \(a_n=\)

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