Determine the independent term of x: (3x−(2/x))^4


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Question Number 72784 by Maclaurin Stickker last updated on 02/Nov/19

$D e t e r m i n e t h e i n d e p e n d e n t t e r m o f$ $x :$ ${(3 x - \frac{2}{x})}^{4}$
Commented by mathmax by abdo last updated on 03/Nov/19

$w e h a v e {(3 x - \frac{2}{x})}^{4} = \sum_{k = 0}^{4} C_{4}^{k} {(3 x)}^{k} {(- \frac{2}{x})}^{4 - k}$ $= \sum_{k 0}^{4} C_{4}^{k} 3^{k} x^{k} {(- 2)}^{4 - k} x^{k - 4} = \sum_{k = 0}^{4} C_{4}^{k} 3^{k} {(- 2)}^{4 - k} x^{2 k - 4}$ $t h e i n d e p e n d e n t t e r m o f x i s o b t a i n e d w h e n 2 k - 4 = 0 \Rightarrow k = 2 \Rightarrow$ $t h e c o e f f i c i e n t i s a_{0} = C_{4}^{2} 3^{2} {(- 2)}^{2} = 36 \times C_{4}^{2}$ $C_{4}^{2} = \frac{4!}{2! (2)!} = \frac{4 \times 3 \times 2!}{2! \times 2!} = 6 \Rightarrow a_{0} = 36 \times 6 = 216$
	Answered by $@ty@m123 last updated on 02/Nov/19

	$t_{r + 1} =^{4} C_{r} {(3 x)}^{4 - r} {(- \frac{2}{x})}^{r}$ $T h e t e r m i s i n d e p e n d e n t o f x i f$ $4 - r = r \Rightarrow 2 r = 4 \Rightarrow r = 2$ $\Rightarrow t_{3} =^{4} C_{2} \times 3^{2} \times {(- 2)}^{2} = 6 \times 9 \times 4 = 216$
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