Given that an = 3an-1 + 1, where a1 = 2, find the 3rd-6th terms of the sequence.A) 2, 5, 11, 34 B) 2, 7, 22, 67 C) 7, 22, 67, 202 D) 22, 67, 202, 607

Answer: D = 22, 67, 202, 607Step-by-step explanation:Given; a₁ = 2aₙ = 3aₙ₋₁ + 1a₂ = 3a₂₋₁ + 1 = 3a₁ + 1 = 3(2) + 1 (recall a₁ = 2) = 6+1 =7a₂ =7a₃ = 3a₃₋₁ + 1 =3a₂ + 1 = 3(7) + 1 (recall a₂=7) =21 + 1 =22a₃ = 22a₄= 3a₄₋₁ =3a₃ + 1 = 3(22) + 1 = 66 + 1 = 67a₄=67a₅ = 3a₅₋₁ =3a₄ + 1 = 3(67) + 1 = 201 + 1 =202a₅=202a₆ =3a₆₋₁ + 1 = 3a₅ + 1 =3(202) + 1 = 606 + 1 =607a₆ =607