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(D) The given function is $f(x) = \frac{1}{(1 - ax)(1 - bx)}$.Using partial fractions,we can write:$\frac{1}{(1 - ax)(1 - bx)} = \frac{A}{1 - ax} + \f

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

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(D) The given function is $f(x) = \frac{1}{(1 - ax)(1 - bx)}$.Using partial fractions,we can write:$\frac{1}{(1 - ax)(1 - bx)} = \frac{A}{1 - ax} + \frac{B}{1 - bx}$.Solving for $A$ and $B$,we get $1 = A(1 - bx) + B(1 - ax)$.For $x = 1/a$,$1 = A(1 - b/a) \implies A = \frac{a}{a - b}$.For $x = 1/b$,$1 = B(1 - a/b) \implies B = \frac{b}{b - a}$.Thus,$f(x) = \frac{a}{a - b}(1 - ax)^{-1} + \frac{b}{b - a}(1 - bx)^{-1}$.Expanding using the binomial series $(1 - z)^{-1} = \sum_{n=0}^{\infty} z^n$:$f(x) = \frac{a}{a - b} \sum_{n=0}^{\infty} (ax)^n + \frac{b}{b - a} \sum_{n=0}^{\infty} (bx)^n$.The coefficient of $x^n$ is $a_n = \frac{a}{a - b} a^n + \frac{b}{b - a} b^n$.$a_n = \frac{a^{n+1}}{a - b} + \frac{b^{n+1}}{b - a} = \frac{b^{n+1} - a^{n+1}}{b - a}$.

If the expansion in powers of $x$ of the function $\frac{1}{(1 - ax)(1 - bx)}$ is $a_0 + a_1x + a_2x^2 + a_3x^3 + \dots$,then $a_n$ is

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