Question

1- O coeficiente de $x^{5}$ no desenvolvimento de $(\frac{2}{x}+x^{3} )^{7}$ é: a) 30 b) 90 c) 120 d) 270 e) 560

Answer 1

Explicação passo-a-passo:

O termo geral é:

$\sf T_{p+1}=\dbinom{n}{p}\cdot \left(\dfrac{2}{x}\right)^{n-p}\cdot(x^3)^{p}$

$\sf T_{p+1}=\dbinom{7}{p}\cdot \left(\dfrac{2}{x}\right)^{7-p}\cdot(x^3)^{p}$

$\sf T_{p+1}=\dbinom{7}{p}\cdot2^{7-p}\cdot \dfrac{1}{x^{7-p}}\cdot x^{3p}$

$\sf T_{p+1}=\dbinom{7}{p}\cdot2^{7-p}\cdot\dfrac{x^{3p}}{x^{7-p}}$

$\sf T_{p+1}=\dbinom{7}{p}\cdot2^{7-p}\cdot x^{3p-(7-p)}$

$\sf T_{p+1}=\dbinom{7}{p}\cdot2^{7-p}\cdot x^{3p-7+p}$

$\sf T_{p+1}=\dbinom{7}{p}\cdot2^{7-p}\cdot x^{4p-7}$

Temos que:

$\sf x^{4p-7}=x^5$

Igualando os expoentes:

$\sf 4p-7=5$

$\sf 4p=5+7$

$\sf 4p=12$

$\sf p=\dfrac{12}{4}$

$\sf p=3$

O termo procurado é:

$\sf T_{3+1}=\dbinom{7}{3}\cdot2^{7-3}\cdot x^{4\cdot3-7}$

$\sf T_4=35\cdot2^4\cdot x^{12-7}$

$\sf T_4=35\cdot16x^5$

$\sf T_4=560x^5$

Letra E