If-and-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-is-equal-to-

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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