Question

determine o termo independente de (x^2+1/x)^12​

Answer 1

Os termos da expansão de (x²+1/x)^{12} podem ser determinados pela formula do termo geral.

$T_{p+1}~=~\binom{n}{p}~.~\left(a^{n-p}~.~b^{\,p}\right)$

O enunciado nos fornece o valor de "n" (12), de "a" (x²) e de "b" (1/x). Para acharmos o valor de "p" vamos utilizar outra informação do enunciado: "termo independente".

O termo independente é o termo em que a variável (x) tem expoente 0, logo queremos que:

$a^{n-p}~.~b^{\,p}~=~x^0\\\\Substituindo~"a"~e~"b":\\\\\left(x^2\right)^{n-p}~.~\left(\frac{1}{x}\right)^p~=~x^0\\\\\\x^{\,2~.~(n-p)}~.~\left(x^{-1}\right)^p~=~x^0\\\\\\x^{2n-2p}~.~x^{-p}~=~x^0\\\\\\x^{2n-2p+(-p)}~=~x^0\\\\\\x^{2n-3p}~=~x^0\\\\\\2n-3p~=~0\\\\\\p~=~\frac{2n}{3}~~~Substituindo~o~valor~de~"n"\\\\\\p~=~\frac{2~.~12}{3}\\\\\\\boxed{p~=~8}$

Agora sim, aplicando a formulação do termo geral:

$T_{p+1}~=~\binom{n}{p}~.~\left(a^{n-p}~.~b^{\,p}\right)\\\\\\T_{8+1}~=~\binom{12}{8}~.~\left(\left(x^2\right)^{12-8}~.~\left(\frac{1}{x}\right)^{\,8}\right)\\\\\\T_{9}~=~\frac{12!}{8!.(12-8)!}~.~\left(\left(x\right)^{2~.~4}~.~\left(x\right)^{-1~.~8}\right)\\\\\\T_{9}~=~\frac{12~.~11~.~10~.~9~.~8!}{8!~.~4!}~.~\left(x^{8}~.~x^{-8}\right)\\\\\\T_{9}~=~\frac{12~.~11~.~10~.~9}{24}~.~\left(x^0\right)\\\\\\T_{9}~=~\frac{11880}{24}~.~x^0\\\\\\\boxed{T_{9}~=~495~.~x^0}$

Resposta: O termo independente vale 495.