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Question Number 124247 by ZiYangLee last updated on 02/Dec/20

Commented by ZiYangLee last updated on 02/Dec/20

$8 and 9, help sir . . .$
	Answered by bramlexs22 last updated on 02/Dec/20

	$(8) ∙ {(x^{2} + \frac{1}{x})}^{n} = \underset{r = 0}{\sum^{n}} (\begin{matrix} n \\ r \end{matrix}) (x^{2 (n - r)}) (x^{- r})$ $T_{4} = (\begin{matrix} n \\ 3 \end{matrix}) x^{2 n - 6} . x^{- 3} = \frac{n . (n - 1) . (n - 2)}{6} x^{9 - n}$ $T_{13} = (\begin{matrix} n \\ 12 \end{matrix}) x^{2 n - 24} . x^{- 12} = \frac{n!}{12! (n - 12)!} . x^{36 - n}$ $c o e f f i c i e n t o f 4^{t h} a n d 13^{t h} a r e e q u a l$ $t h e n \frac{n!}{3! (n - 3)!} = \frac{n!}{12! (n - 12)!}$ $\Leftrightarrow 3! (n - 3) (n - 4) . . . (n - 12)! = 12.11 . . . 3! (n - 12)!$ $w e g e t n = 15 . r e c a l l f o r m u l a$ $(\begin{matrix} n \\ r \end{matrix}) = (\begin{matrix} n \\ n - r \end{matrix})$ $w e w a n t t o f i n d t h e t e r m i n d e n p e n d e n t$ $o f x i n t h e e x p a n s i o n {(x^{2} + \frac{1}{x})}^{15}$ ${(x^{2} + \frac{1}{x})}^{15} = \underset{r = 0}{\sum^{15}} (\begin{matrix} 15 \\ r \end{matrix}) x^{30 - 2 r} . (x^{- r})$ $t h e t e r m i n d e p e n d e n t o f x w e g e t i f$ $x^{30 - 3 r} = x^{0}; r = 10 . T h u s t h e t e r m$ $i n d e p e n d e n t o f x = (\begin{matrix} 15 \\ 10 \end{matrix}) = \frac{15.14.13.12.11}{5.4.3.2.1} = 3003$ $\frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}$ $3003.0$
	Commented by ZiYangLee last updated on 02/Dec/20

	$wow thanks!$
	Answered by bramlexs22 last updated on 02/Dec/20

	$(9) ∙ ∙ {(1 + x)}^{18} = \underset{r = 0}{\sum^{18}} (\begin{matrix} 18 \\ r \end{matrix}) x^{18 - r}$ $c o e f f i c i e n t o f t h e {(2 r + 4)}^{t h} = (\begin{matrix} 18 \\ 2 r + 3 \end{matrix}) = \frac{18!}{(2 r + 3)! (15 - 2 r)!}$ $c o e f f i c i e n t o f {(r - 2)}^{t h} = (\begin{matrix} 18 \\ r - 3 \end{matrix}) = \frac{18!}{(r - 3)! (21 - r)!}$ $t h e (2 r + 3)! (15 - 2 r)! = (r - 3)! (21 - r)!$ $w e g e t r = 6 .$
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