If frac{(1-px)^{-1}}{(1-qx)}=a_0+a_1x+a_2x^2+ | vedclass

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(B) Given the expression: $\frac{(1-px)^{-1}}{(1-qx)} = a_0+a_1x+a_2x^2+\ldots+a_nx^n+\ldots$ Note that the expression is equivalent to $(1-px)^{-1}(1

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$\frac{p^{n+1}-q^{n+1}}{p-q}$

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(B) Given the expression: $\frac{(1-px)^{-1}}{(1-qx)} = a_0+a_1x+a_2x^2+\ldots+a_nx^n+\ldots$ Note that the expression is equivalent to $(1-px)^{-1}(1-qx)^{-1}$ only if the denominator was $(1-qx)^{-1}$. However,the problem states $\frac{(1-px)^{-1}}{(1-qx)}$. Wait,the standard expansion for $\frac{1}{(1-px)(1-qx)}$ is $\sum a_n x^n$. Using partial fractions: $\frac{1}{(1-px)(1-qx)} = \frac{A}{1-px} + \frac{B}{1-qx}$. $1 = A(1-qx) + B(1-px)$. For $x = 1/p$,$1 = A(1-q/p) \implies A = \frac{p}{p-q}$. For $x = 1/q$,$1 = B(1-p/q) \implies B = \frac{q}{q-p} = -\frac{q}{p-q}$. So,$\frac{1}{(1-px)(1-qx)} = \frac{p}{p-q} \sum (px)^n - \frac{q}{p-q} \sum (qx)^n$. $a_n = \frac{p}{p-q} p^n - \frac{q}{p-q} q^n = \frac{p^{n+1}-q^{n+1}}{p-q}$. Thus,$a_n = \frac{p^{n+1}-q^{n+1}}{p-q}$.

If $\frac{(1-px)^{-1}}{(1-qx)}=a_0+a_1x+a_2x^2+a_3x^3+\ldots$,then $a_n=$

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