Question

Determine os quatros primeiros termos das sequências definidas por:
A: An=n2+2
B: An=1-3n
C: An=3n+2/n+1

Answer 1

Resolução da questão, veja:

Letra “A”:

$\mathsf{A_{n} = n^{2} + 2}}~\to~ \mathsf{A_{1} = 1.}}\\\\\\ \mathsf{A_{1} = 1^{2} + 2}\\\\\ \mathsf{A_{1} = 3.}}}}}}}\\\\\\\\ \mathsf{A_{n} = n^{2} + 2}}~\to~\mathsf{A_{2} = 2.}}\\\\\\ \mathsf{A_{2} = 2^{2} + 2}\\\\\ \mathsf{A_{2} = 4.}}}}}}}\\\\\\\\ \mathsf{A_{n} = n^{2} + 2}}~\to~\mathsf{A_{3} = 3.}}\\\\\\ \mathsf{A_{3} = 3^{2} + 2}\\\\\ \mathsf{A_{3} = 11.}}}}}}}\\\\\\\\ \mathsf{A_{n} = n^{2} + 2}}~\to~\mathsf{A_{4} = 4.}}\\\\\\ \mathsf{A_{4} = 4^{2} + 2}\\\\\ \mathsf{A_{4} = 18.}}}}}}}$

Letra “B”:

$\mathsf{A_{n} = 1-3n}}~\to~\mathsf{A_{1} = 1}}\\\\\\ \mathsf{A_{1} = 1 - 3~\cdot~1}\\\\\ \mathsf{A_{1} = 1 - 3}}\\\\\\ \mathsf{A_{1} = -2.}}\\\\\\\ \mathsf{A_{n} = 1-3n}}~\to~\mathsf{A_{2} = 2.}}\\\\\\ \mathsf{A_{2} = 1 - 3~\cdot~2}\\\\\ \mathsf{A_{2} = 1 - 6}}\\\\\\ \mathsf{A_{2} = -5.}}\\\\\\\ \mathsf{A_{n} = 1-3n}}~\to~\mathsf{A_{3} = 3.}}\\\\\\ \mathsf{A_{3} = 1 - 3~\cdot~3}\\\\\ \mathsf{A_{3} = 1 - 9}}\\\\\\ \mathsf{A_{3} = -8.}}\\\\\\\ \mathsf{A_{n} = 1-3n}}~\to~\mathsf{A_{4} = 4.}}\\\\\\ \mathsf{A_{4} = 1 - 3~\cdot~4}\\\\\ \mathsf{A_{4} = 1 - 12}}\\\\\\ \mathsf{A_{4} = -11.}}$

Letra “C”:

$\mathsf{A_{n} = \dfrac{3n + 2}{n+1}}~\to~ \mathsf{A_{1} = 1.}}\\\\\\\ \mathsf{A_{1} = \dfrac{3~\cdot~1+2}{1+1}}}\\\\\\\\ \mathsf{A_{1} = \dfrac{5}{2}}}\\\\\\\ \mathsf{A_{n} = \dfrac{3n + 2}{n+1}}~\to~ \mathsf{A_{2} = 2.}}}\\\\\\\\ \mathsf{A_{2} = \dfrac{3~\cdot~2+2}{2+1}}\\\\\\\\ \mathsf{A_{2} = \dfrac{8}{3}}\\\\\\\ \mathsf{A_{n} = \dfrac{3n + 2}{n+1}}~\to~ \mathsf{A_{3} = 3.}}}\\\\\\\ \mathsf{A_{3} = \dfrac{3~\cdot~3+2}{3+1}}\\\\\\\ \mathsf{A_{3} = \dfrac{11}{4}}\\\\\\\\ \mathsf{A_{n} = \dfrac{3n + 2}{n+1}}~\to~ \mathsf{A_{4} = 4.}}\\\\\\\ \mathsf{A_{4} = \dfrac{3~\cdot~4+2}{4+1}}\\\\\\\ \mathsf{A_{4} = \dfrac{14}{5}}$

Espero que te ajude '-'