Recursion formula

Question

a 1 = 4 and a n = 3a n-1 -3 for n > (greater than or equl to ) 2

write first 4 terms

Answer 1

a₁=4

a_n=3a_n-1-3

2≤n

a₂=3a_2-1-3= 3a₁-3 = 3·4-3= 9

a₂=9

a₃=3a_3-1-3= 3a₂-3 = 3·9-3= 24

a₃=24

a₄=3a_4-1-3= 3a₃-3 = 3·24-3= 69

a₄=69