Question -
Answer -
Given:
a1 = 3and an = 3an–1 + 2, for all n > 1
By using the values n= 1, 2, 3, 4 we can find the first four terms.
When n = 1:
a1 = 3
When n = 2:
a2 =3a2–1 + 2
= 3a1 +2
= 3(3) + 2
= 9 + 2
= 11
When n = 3:
a3 =3a3–1 + 2
= 3a2 +2
= 3(11) + 2
= 33 + 2
= 35
When n = 4:
a4 =3a4–1 + 2
= 3a3 +2
= 3(35) + 2
= 105 + 2
= 107
∴ First four termsof sequence are 3, 11, 35, 107.