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

D = 22, 67, 202, 607

Step-by-step explanation:

Given;

a₁ = 2

aₙ = 3aₙ₋₁ + 1

a₂ = 3a₂₋₁ + 1

= 3a₁ + 1

= 3(2) + 1 (recall a₁ = 2)

= 6+1

=7

a₂ =7

a₃ = 3a₃₋₁ + 1

=3a₂ + 1

= 3(7) + 1 (recall a₂=7)

=21 + 1

=22

a₃ = 22

a₄= 3a₄₋₁

=3a₃ + 1

= 3(22) + 1

= 66 + 1

= 67

a₄=67

a₅ = 3a₅₋₁

=3a₄ + 1

= 3(67) + 1

= 201 + 1

=202

a₅=202

a₆ =3a₆₋₁ + 1

= 3a₅ + 1

=3(202) + 1

= 606 + 1

=607

a₆ =607

(2,6)

step-by-step explanation:

cause i just did this and got the answer

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