[tex]\boxed{T_{p+1} = \begin{pmatrix}
n \\
p
\end{pmatrix} \cdot a^{p} \cdot x^{n-p}}[/tex]
Pelas informações, temos que:
n = 18 (potência do termo)
p = temos que descobrir
a = 1/x^{2}
x = raiz 4 de x
Substituindo:
[tex]T_{p+1} = \begin{pmatrix} n \\ p \end{pmatrix} \cdot a^{p} \cdot x^{n-p}
\\\\\\
T_{p+1} = \begin{pmatrix} 18 \\ p \end{pmatrix} \cdot (\frac{1}{x^{2}})^{p} \cdot (x^{\frac{1}{4}})^{18-p}
\\\\\\
T_{p+1} = \begin{pmatrix} 18 \\ p \end{pmatrix} \cdot (x^{2}})^{-p} \cdot x^{\frac{18}{4}-\frac{p}{4}}
\\\\\\
T_{p+1} = \begin{pmatrix} 18 \\ p \end{pmatrix} \cdot x^{-2p} \cdot x^{\frac{18}{4}-\frac{p}{4}}[/tex]
Quando há multiplicação de bases iguais, somamos:
[tex]x^{-2p} \cdot x^{\frac{18}{4}-\frac{p}{4}[/tex]
[tex]-2p+\frac{18}{4}-\frac{p}{4}
\\\\
\frac{-2p^{\times 4}}{1^{\times 4}} + \frac{18}{4}-\frac{p}{4}
\\\\
\frac{-8p}{4} + \frac{18}{4}-\frac{p}{4} = \boxed{\frac{18}{4}-\frac{9p}{4}}[/tex]
[tex]T_{p+1} = \begin{pmatrix} 18 \\ p \end{pmatrix} \cdot x^{\frac{18}{4}-\frac{9p}{4}}[/tex]
Agora igualamos a x elevado a 0, pois todo número elevado a zero é 1. E como quer o termo independente, não podemos ter incógnita junto.
[tex]\not{x}^{\frac{18}{4}-\frac{9p}{4} =\not{x}^{0}[/tex]
[tex]\frac{18}{\not{4}}-\frac{9p}{\not{4}} =0
\\\\\
9p = 18
\\\\
p = \frac{18}{9}
\\\\
\boxed{p = 2}[/tex]
Agora voltamos para substituir:
[tex]T_{p+1} = \begin{pmatrix} 18 \\ p \end{pmatrix} \cdot x^{\frac{18}{4}-\frac{9p}{4}}
\\\\\\
T_{2+1} = \begin{pmatrix} 18 \\ 2 \end{pmatrix} \cdot x^{\frac{18}{4}-\frac{9 \cdot (2)}{4}}
\\\\\\
T_{3} = \begin{pmatrix} 18 \\ 2 \end{pmatrix} \cdot x^{\frac{18}{4}-\frac{18}{4}}
\\\\\\
T_{3} = \begin{pmatrix} 18 \\ 2 \end{pmatrix} \cdot x^{0}
\\\\\\
T_{3} = \begin{pmatrix} 18 \\ 2 \end{pmatrix} \cdot 1[/tex]
É só resolvermos a o fatorial:
[tex]\frac{n!}{p! \cdot (n-p)!}
\\\\
\frac{18!}{2! \cdot (18-2)!}
\\\\
\frac{18!}{2! \cdot (18-2)!}
\\\\
\frac{18!}{2! \cdot 16!}
\\\\
\frac{18 \cdot 17 \cdot \not{16!}}{2 \cdot \not{16!}} = \frac{306}{2} = 153[/tex]
[tex]T_{3} = \begin{pmatrix} 18 \\ 2 \end{pmatrix}
\\\\\\
\boxed{\boxed{\boxed{T_{3} = 153}}} \rightarrow termo \ independente[/tex]
