[tex]\displaystyle \underline{\text{Bin{\^o}mio}}}: \\\\\ (\text A+\text B)^{\text n}=\sum_{\text k\ =\ 0}^{\text n}{\text n\choose \text k }\text{A}^{(\text n-\text k)}.\text B^{\text k} \\\\\\\ \underline{\text{Posi{\c c}{\~a}o do termo}}: \\\\ \text T_{\text k+1}={\text n\choose \text k }\text{A}^{(\text n-\text k)}.\text B^{\text k}[/tex]
Item a)
O quarto termo de [tex](4\text x+1)^7[/tex]
[tex]\displaystyle \text{T}_\text{k+1} = {7\choose \text k}(4\text x)^{(7-\text k)}.1^{\text k}[/tex]
[tex]\text T_\text{k+1=}\text T_\text{4} \to \to \text k+1=4 \to \boxed{\text k =3}[/tex]
Daí :
[tex]\displaystyle \text T_4={7\choose 3}(4\text x)^{(7-3)}.1^3 \\\\\\\ \text T_4=\frac{7!}{(7-3)!3!}(4\text x)^{4} \\\\\\ \text T_4=\frac{7.6.5.4!}{4!3! }(4^4\text x^4) \\\\\\ \text T_4=\frac{7.6.5}{3.2}(2^8\text x^4) \\\\\\ \text T_4=35.256.\text x^4 \\\\ \huge\boxed{\text T_4=8960.\text x^4\ }\checkmark[/tex]
item b)
O sétimo termo de [tex](\text x-1)^{10}[/tex]
[tex]\displaystyle \text T_\text{k+1}=\text T_7 \to \text k+1=7 \to \boxed{\text k =6} \\\\\\ \text T_{\text k+1}={10\choose \text k}.\text x^{10- \text k}(-1)^{\text k} \\\\\\\ \text T_{7}={10\choose 6}\text x^{(10-6)}(-1)^{6} \\\\\\ \text T_6=\frac{10!}{(10-6)!6!}.\text x^4 \\\\\\ \text T_7= \frac{10.9.8.7.6!}{4!6!}.\text x^4 \\\\\\ \text T_7=\frac{10.9.8.7}{4.3.2}.\text x^4 \\\\\\ \huge\boxed{\text T_7=210.\text x^4\ }\checkmark[/tex]
