coeff of x^2 from [6+8x−(3/2)x^2 ]^5


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Question Number 176276 by cortano1 last updated on 15/Sep/22

$coeff of x^{2} from$ ${[6 + 8 x - \frac{3}{2} x^{2}]}^{5}$
	Answered by mr W last updated on 15/Sep/22

	$C_{1}^{5} \times (- \frac{3}{2}) \times 6^{4} + C_{2}^{5} \times 8^{2} \times 6^{3} = 128520$
	Commented by cortano1 last updated on 16/Sep/22

	$nice .$
	Commented by cortano1 last updated on 16/Sep/22

	$it is from form$ $\underset{k = 0}{\sum^{5}} (\begin{matrix} 5 \\ k \end{matrix}) {(6 + 8 x)}^{5 - k} {(- \frac{3}{2} x^{2})}^{k} ?$
	Commented by mr W last updated on 16/Sep/22

	$n o, i a p p l i e d n o f o r m u l a .$ $t o g e t x^{2} w e n e e d$ $1 . o n e t e r m - \frac{3}{2} x^{2} a n d 4 t e r m s 6,$ $t o s e l e c t t h e t e r m - \frac{3}{2} x^{2}, t h e r e a r e$ $C_{1}^{5} w a y s, t h e r e f o r e w e g e t$ $C_{1}^{5} \times (- \frac{3}{2}) \times 6^{4} .$ $2 . t w o t e r m s 8 x a n d 3 t e r m s 6,$ $t o s e l e c t t w o t e r m s 8 x, t h e r e a r e$ $C_{2}^{5} w a y s, t h e r e f o r e w e g e t$ $C_{2}^{5} \times 8^{2} \times 6^{3} .$ $t o t a l l y w e g e t$ $C_{1}^{5} \times (- \frac{3}{2}) \times 6^{4} + C_{2}^{5} \times 8^{2} \times 6^{3}$
	Commented by Tawa11 last updated on 18/Sep/22

	$Great sir$
	Answered by mr W last updated on 16/Sep/22

	${(6 + 8 x - \frac{3}{2} x^{2})}^{5}$ $= \underset{0 ⩽ a, b, c ⩽ 5}{\sum^{a + b + c = 5}} (\begin{matrix} 5 \\ a, b, c \end{matrix}) \times 6^{a} \times {(8 x)}^{b} \times {(- \frac{3}{2} x^{2})}^{c}$ $t o g e t x^{2} t e r m : b + 2 c = 2$ $\Rightarrow b = 0, c = 1, a = 4 o r$ $\Rightarrow b = 2, c = 0, a = 3 .$ $t h e c o e f f . o f x^{2} t e r m i s :$ $(\begin{matrix} 5 \\ 4, 0, 1 \end{matrix}) \times 6^{4} \times {(8)}^{0} \times {(- \frac{3}{2})}^{1} + (\begin{matrix} 5 \\ 3, 2, 0 \end{matrix}) \times 6^{3} \times {(8)}^{2} \times {(- \frac{3}{2})}^{0}$ $= \frac{5!}{4! 0! 1!} \times 6^{4} \times {(8)}^{0} \times {(- \frac{3}{2})}^{1} + \frac{5!}{3! 2! 0!} \times 6^{3} \times {(8)}^{2} \times {(- \frac{3}{2})}^{0}$ $= 128520$
	Commented by Tawa11 last updated on 25/Sep/22

	$Great sir .$
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