Question

Determine o coeficiente de x⁸ no desenvolvimento de​

Answer 1

Resposta:

$(a+b)^{n}$

$T_{p+1} = C_{n,p} * a^{n-p} * b^{p}$

(x⁴)^(n-p) * (-1/x)^p =x⁸

para n=7

(x⁴)^(7-p) * (-1/x)^p =x⁸

x^(28-4p-p) *(-1)^p=x⁸

p tem que ser par

(28-4p-p)=8   ==> p = 4

C7,4 =7!/(7-4)!4!  = 7!/3!4!= 35

Letra C

#(x⁴-1/x)⁷ =   -1/x^7 + 7/x^2 - 21 x^3 + 35 x^8 - 35 x^13 + 21 x^18 - 7 x^23 + x^28

Answer 2

Resposta:

c) 35

Explicação passo a passo:

$T_{geral} = \binom{n}{p} (x^4)^p . (\frac{-1}{x} )^{n-p}\\\\T_{x^8} = \binom{7}{p} (x^4)^p . (\frac{-1}{x} )^{7-p}\\\\T_{x^8} = \binom{7}{p} x^{4p} . \frac{(-1)^{7-p}}{x^{7-p}}\\\\T_{x^8} = \binom{7}{p} \frac{x^{4p}}{x^{7-p}} . (-1)^{7-p}\\\\T_{x^8} = \binom{7}{p} x^{4p-(7-p)} . (-1)^{7-p}\\\\4p - (7-p) = 8\\\\4p - 7 + p = 8\\\\5p = 15\\\\p = 3\\\\T_{x^8} = \binom{7}{3} x^{4.3-(7-3)} . (-1)^{7-3}\\\\T_{x^8} = \binom{7}{3} x^{8} . (-1)^{4}\\\\T_{x^8} = 35 x^{8}$