Question

Complete os 5 primeiros termos da sequencia sendo n a posição de cada termo.a)3*nb)2*n-2
c)2n-1
d)n²-1
e)2.(n+1)

Answer 1

$a) a_{n} = 3n \\ \\ a_{1} =3(1) --> a_{1} =3 \\ a_{2} = 3(2)--> a_{2} =6 \\ a_{3} =3(3)--> a_{3} =9 \\ a_{4} =3(4)--> a_{4} = 12 \\ a_{5} =3(5)--> a_{5} =15 \\ \\ S=( a_{1}, a_{2} , a_{3} , a_{4}, a_{5}) \\ S=(3, 6, 9, 12, 15)$


$b) a_{n}=2n-2 \\ \\ a_{1}=2(1)-2--> a_{1}=0 \\ a_{2}=2(2)-2--> a_{2}=2 \\ a_{3}=2(3)-2--> a_{3}=4 \\ a_{4}=2(4)-2--> a_{4}=6 \\ a_{5}=2(5)-2--> a_{5}=8 \\ \\ S=(0,2,4,6,8)$


$c) a_{n}=2n-1 \\ \\ a_{1}=2(1)-1--> a_{1}=1 \\ a_{2}=2(2)-1--> a_{2}=3 \\ a_{3}=2(3)-1--> a_{3}=5 \\ a_{4}=2(4)-1--> a_{4}=7 \\ a_{5}=2(5)-1--> a_{5}=9 \\ \\ S=(1,3,5,7,9)$


$d)a_{n}= n^{2}-1 \\ \\ a_{1}= 1^{2}-1--> a_{1} =0 \\ a_{2}= 2^{2}-1--> a_{2}=3 \\ a_{3}= 3^{2}-1--> a_{3}=8 \\ a_{4}= 4^{2}-1-->a_{4}=15 \\ a_{5}= 5^{2}-1-->a_{5}=24 \\ \\ S=(1,3,8,15,24)$


$e)a_{n}=2(n+1) \\ \\ a_{1}=2(1+1)-->a_{1}=4 \\ a_{2}=2(2+1)-->a_{2}=6 \\ a_{3}=2(3+1)-->a_{3}=8 \\ a_{4}=2(4+1)-->a_{4}=10 \\ a_{5}=2(5+1)--> a_5=12 \\ \\ S=(4,6,8,10,12) \\ \\ \\ Renato.$