Question

o termo independente de x do desenvolvimento de
$(x + \frac{1}{ {x}^{3} } {)}^{12}$
  é ?​

Answer 1

Explicação passo-a-passo:

Binómio de Newton

Dado o binómio :

$\sf{ \red{ \Big( x + \dfrac{1}{x^3} \Big)^{12} } }$

Usando a fórmula do termo geral binomial :

$\sf{ \huge{ T_{p + 1}~=~ C^{n}_{p} * a^{n - p}*b^{p} } }$

Então :

$\iff \sf{ T_{p + 1}~=~ C^{12}_{p}* x^{12-p}*(x^{-3})^{p} }$

$\iff \sf{ T_{p + 1}~=~ C^{12}_{p} * x^{12-p}*x^{-3p} }$

$\iff \sf{ T_{p + 1}~=~ C^{12}_{p} * x^{12 - 4p} }$

Para que este termo seja independente é necessário que o expoente em x seja igual a zero.

$\blue{ \iff \sf{ 12 - 4p~=~ 0 }}$

$\iff \sf{ 4p~=~12 }$

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

$\iff \boxed{ \sf{ \pink{ p~=~ 3 } }}$

Logo vamos ter que :

$\iff \sf{ T_{3 + 1}~=~ C^{12}_{4} * x^{0} }$

$\iff \sf{ T_{4}~=~ \dfrac{12!}{3!(12 - 3)!} * 1 }$

$\iff \sf{ T_{4}~=~ \dfrac{\cancel{12}*11*10*\cancel{9!}}{\cancel{6}* \cancel{9!} } }$

$\iff \sf{ T_{4}~=~ 2*11*10~=~22*10 }$

$\green{ \iff \boxed{ \boxed{ \sf{ T_{4}~=~220 } } } \sf{ \longleftarrow Resposta } }$

Espero ter ajudado bastante!)

Answer 2

Explicação passo-a-passo:

$\sf \left(x+\dfrac{1}{x^3}\right)^{12}$

O termo geral é dado por:

$\sf T_{p+1}=\dbinom{12}{p}\cdot x^{12-p}\cdot\left(\dfrac{1}{x^3}\right)^p$

$\sf T_{p+1}=\dbinom{12}{p}\cdot x^{12-p}\cdot\dfrac{1}{x^{3p}}$

$\sf T_{p+1}=\dbinom{12}{p}\cdot\dfrac{x^{12-p}}{x^{3p}}$

$\sf T_{p+1}=\dbinom{12}{p}\cdot x^{12-p-3p}$

$\sf T_{p+1}=\dbinom{12}{p}\cdot x^{12-4p}$

Temos que:

$\sf x^{12-4p}=x^0$

Igualando os expoentes:

$\sf 12-4p=0$

$\sf 4p=12$

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

$\sf p=3$

Logo, o termo independente de x é:

$\sf T_{3+1}=\dbinom{12}{3}\cdot x^{12-4\cdot3}$

$\sf T_{4}=\dfrac{12\cdot11\cdot10}{3!}\cdot x^{12-12}$

$\sf T_{4}=\dfrac{1320}{6}\cdot x^0$

$\sf T_{4}=220\cdot1$

$\sf \red{T_{4}=220}$