The coefficient of x 18 in the expansion of ( x 4 - 1 x 3 ) 15 is

Question

The coefficient of $x^{18}$ in the expansion of ${(x^{4} - \frac{1}{x^{3}})}^{15}$ is _________.

Smart Achievers · Accepted Answer

(5005)

$Given,$ ${(x^{4} - \frac{1}{x^{3}})}^{15}$

$T_{r + 1} = {}^{15}C_{r} {(x^{4})}^{15 - r} {(- \frac{1}{x^{3}})}^{r}$

$= {}^{15}C_{r} {(x)}^{60 - 4 r} {\cdot (- 1)}^{r} x^{- 3 r} = {}^{15}C_{r} {(x)}^{60 - 7 r} \cdot {(- 1)}^{r}$

$For coefficient of x^{18}, put 60 - 7 r = 18 \Rightarrow r = \frac{42}{7} = 6$

$Hence, coefficient of x^{18} is {}^{15}C_{6} {(- 1)}^{6} = 5005$

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