Question

1- Calcule os 5 primeiros termos das sequências abaixo:

1 - an = 4n +1
2- an = 5n² -1
3- an = n² + 1
4- an = 3n² + n
5- an = 2n²
6- an = n² + n
7- an = - n² + 1
8- an = - 3n² -3
9- an = - 2n² -1
10- an - 3 - n²

Answer 1

Resposta:

$\text{Leia abaixo}$

Explicação passo-a-passo:

$1)\: a_n = 4n + 1$

$a_1 = 4(1) + 1 = 5\\a_2 = 4(2) + 1 = 9\\a_3 = 4(3) + 1 = 13\\a_4 = 4(4) + 1 = 17\\a_5 = 4(5) + 1 = 21$

$2)\: a_n = 5n^2 - 1$

$a_1 = 5(1)^2 - 1 = 4\\a_2 = 5(2)^2 - 1 = 19\\a_3 = 5(3)^2 - 1 = 44\\a_4 = 5(4)^2 - 1 = 79\\a_5 = 5(5)^2 - 1 = 124$

$3)\: a_n = n^2 + 1$

$a_1 = (1)^2 + 1 = 2\\a_2 = (2)^2 + 1 = 5\\a_3 = (3)^2 + 1 = 10\\a_4 = (4)^2 + 1 = 17\\a_5 = (5)^2 + 1 = 26$

$4)\: a_n = 3n^2 + n$

$a_1 = 3(1)^2 + 1 = 4\\a_2 = 3(2)^2 + 2 = 14\\a_3 = 3(3)^2 + 3 = 30\\a_4 =3(4)^2 + 4 = 52\\a_5 = 3(5)^2 + 5 = 80$

$5)\: a_n = 2n^2$

$a_1 = 2(1)^2 = 2\\a_2 = 2(2)^2 = 8\\a_3 = 2(3)^2 = 18\\a_4 =2(4)^2 = 32\\a_5 = 2(5)^2 = 50$

$6)\: a_n = n^2 + n$

$a_1 = (1)^2 + 1 = 2\\a_2 = (2)^2 + 2 = 6\\a_3 = (3)^2 + 3 = 12\\a_4 =(4)^2 + 4 = 20\\a_5 = (5)^2 + 5 = 30$

$7)\: a_n = -n^2 + 1$

$a_1 = -(1)^2 + 1 = 0\\a_2 = -(2)^2 + 1 = -3\\a_3 = -(3)^2 + 1 = -8\\a_4 = -(4)^2 + 1 = -15\\a_5 = -(5)^2 + 1 = -24$

$8)\: a_n = -3n^2 - 3$

$a_1 = -3(1)^2 - 3 = -6\\a_2 = -3(2)^2 - 3 = -15\\a_3 = -3(3)^2 - 3 = -30\\a_4 =-3(4)^2 - 3 = -51\\a_5 = -3(5)^2 - 3 = -78$

$9)\: a_n = -2n^2 - 1$

$a_1 = -2(1)^2 - 1 = -3\\a_2 = -2(2)^2 - 1 = -13\\a_3 = -2(3)^2 - 1 = -19\\a_4 = -2(4)^2 - 1 = -33\\a_5 = -2(5)^2 - 1 = -51$

$10)\: a_n = -3 - n^2$

$a_1 = -3 - (1)^2 = -4\\a_2 = -3 - (2)^2 = -7\\a_3 = -3 - (3)^2 = -12\\a_4 = -3 - (4)^2 = -19\\a_5 = -3 - (5)^2 = -28$