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For $|x| < \frac{1}{5}$, the coefficient of $x^3$ in the expansion of $\frac{1}{(1-5 x)(1-4 x)}$ is
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The correct answer is:
369
$|x| < \frac{1}{5}$ and $\frac{1}{(1-5 x)(1-4 x)}$
$=(1-5 x)^{-1} \cdot(1-4 x)^{-1}$
then by binomial expansion
$=\left\{1+5 x+25 x^2+125 x^3+\ldots\right\}$ $\cdot\left\{1+4 x+16 x^2+64 x^3+\ldots\right\}$
the coefficient of $x^3$ in this expansion is
$=(125+100+80+64)$
$=369$
$=(1-5 x)^{-1} \cdot(1-4 x)^{-1}$
then by binomial expansion
$=\left\{1+5 x+25 x^2+125 x^3+\ldots\right\}$ $\cdot\left\{1+4 x+16 x^2+64 x^3+\ldots\right\}$
the coefficient of $x^3$ in this expansion is
$=(125+100+80+64)$
$=369$
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