If α and β are the coefficient of x^8 and x^(−24) respectively in the expansion of [ x^4 +2+(1/x^4 ) ]^(10) in powers of x then (α/β) is equal to


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Question Number 131248 by bramlexs22 last updated on 03/Feb/21

$I f α a n d β a r e t h e c o e f f i c i e n t$ $o f x^{8} a n d x^{- 24} r e s p e c t i v e l y$ $i n t h e e x p a n s i o n o f {[x^{4} + 2 + \frac{1}{x^{4}}]}^{10}$ $i n p o w e r s o f x t h e n \frac{α}{β} i s e q u a l t o$
	Answered by EDWIN88 last updated on 03/Feb/21

	${[(x^{4} + x^{- 4}) + 2]}^{10} = \underset{k = 0}{\sum^{20}} (\begin{matrix} 20 \\ k \end{matrix}) {(x^{4} + x^{- 4})}^{20 - k} {(2)}^{k}$ $= \underset{k = 0}{\sum^{20}} (\begin{matrix} 20 \\ k \end{matrix}) 2^{k} [\underset{k = 0}{\sum^{20 - k}} (\begin{matrix} 20 - k \\ k \end{matrix}) {(x^{4})}^{20 - k} {(x^{- 4})}^{k}]$
	Answered by mr W last updated on 03/Feb/21

	${[x^{4} + 2 + \frac{1}{x^{4}}]}^{10}$ $= \frac{1}{x^{40}} {[x^{8} + 2 x^{4} + 1]}^{10}$ $= \frac{1}{x^{40}} {(1 + x^{4})}^{20}$ $= \frac{1}{x^{40}} \underset{k = 0}{\sum^{20}} C_{k}^{20} x^{4 k}$ $= \underset{k = 0}{\sum^{20}} C_{k}^{20} x^{4 k - 40}$ $t e r m x^{8} : 4 k - 40 = 8 \Rightarrow k = 12 \Rightarrow α = C_{12}^{20}$ $t e r m x^{- 24} : 4 k - 40 = - 24 \Rightarrow k = 4 \Rightarrow β = C_{4}^{20}$ $\frac{α}{β} = \frac{C_{12}^{20}}{C_{4}^{20}} = \frac{20!}{12! 8!} \times \frac{4! \times 16!}{20!} = \frac{16 \times 15 \times 14 \times 13}{8 \times 7 \times 6 \times 5}$ $= 26$
	Commented by EDWIN88 last updated on 03/Feb/21

	$n i c e$
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