f(n) = f(n-1) + 3
substitute n= 2 in the above
f(2) = f (2-1) + 3
= f(1) + 3
= 5 + 3
= 8
substitute n = 3 in the formula
f(3) = f(3-1) + 3
= f(2) + 3
= 8 + 3
= 11
substituite n = 4
f(4) = f(4-1) + 3
= f(3) + 3
= 11 + 3
= 14
The first four terms are 5, 8, 11 and 14