Question

No desenvolvimento do binômio (x + a/x)6 , o coeficiente do termo em x4 é 12. Qual o valor de a?

Answer 1

$\boxed{(a+b)^n=\sum\limits_{p=0}^{n}\binom{n}{p}a^{n-p}b^p}$

$\boxed{T_{p+1}=\binom{n}{p}a^{n-p}b^p}$

O binômio é:

$\bigg(x+\dfrac{a}{x}\bigg)^6$

Seu termo geral será:

$T_{p+1}=\dbinom{6}{p}x^{6-p}\bigg(\dfrac{a}{x}\bigg)^p\ \therefore\ \boxed{T_{p+1}=\dbinom{6}{p}x^{2(3-p)}a^p}$

Para o termo $x^4$, o valor de $p$ será:

$2(3-p)=4\ \therefore\ 3-p=2\ \therefore\ \boxed{p=1}$

Dessa forma:

$T_{1+1}=\dbinom{6}{1}x^{2(3-1)}a^1\ \therefore\ T_2=\dfrac{6!}{1!5!}x^4a\ \therefore\ T_2=6ax^4$

Mas $T_2=12x^4$, logo:

$12x^4=6ax^4\ \therefore\ 12=6a\ \therefore\ \boxed{a=2}$

Resposta: o valor de a é 2.