MATH SOLVE

3 months ago

Q:
# 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

Accepted Solution

A:

Answer: D = 22, 67, 202, 607Step-by-step explanation:Given; aβ = 2aβ = 3aβββ + 1aβ = 3aβββ + 1 = 3aβ + 1 = 3(2) + 1 (recall aβ = 2) = 6+1 =7aβ =7aβ = 3aβββ + 1 =3aβ + 1 = 3(7) + 1 (recall aβ=7) =21 + 1 =22aβ = 22aβ= 3aβββ =3aβ + 1 = 3(22) + 1 = 66 + 1 = 67aβ=67aβ
= 3aβ
ββ =3aβ + 1 = 3(67) + 1 = 201 + 1 =202aβ
=202aβ =3aβββ + 1 = 3aβ
+ 1 =3(202) + 1 = 606 + 1 =607aβ =607