Question

Escreva os 6 primeiros termos da sequência 2n²+8:

Answer 1

Resposta:

Explicação passo a passo:

$a_{n}=2n^{2} +8$

$a_{1}=2.(1)^{2} +8=>a_{1} =2.1+8=>a_{1}=2+8=>a_{1} =10$

$a_{2}=2.(2)^{2} +8=>a_{2} =2.4+8=>a_{2}=8+8=>a_{2} =16$

$a_{3}=2.(3)^{2} +8=>a_{3} =2.9+8=>a_{3}=18+8=>a_{3} =26$

$a_{4}=2.(4)^{2} +8=>a_{4} =2.16+8=>a_{4}=32+8=>a_{4} =40$

$a_{5}=2.(5)^{2} +8=>a_{5} =2.25+8=>a_{5}=50+8=>a_{5} =58$

$a_{6}=2.(6)^{2} +8=>a_{6} =2.36+8=>a_{6}=72+8=>a_{6} =80$

   $Resposta:\:(10;16;26;40;58;80)$