If f(n) is a function such that f(1)=f(2)=f(3)=1, and is defined as
f(n+1)=f(n−2)f(n)⋅f(n−1)+1 for n>3,
then f(6)=
2
3
7
11
26
Solution
We will compute f(4),f(5), and f(6):
f(4)=f(3+1)=f(1)f(3)⋅f(2)+1=11⋅1+1=2
f(5)=f(4+1)=f(2)f(4)⋅f(3)+1=12⋅1+1=3
f(6)=f(5+1)=f(3)f(5)⋅f4)+1=13⋅2+1=7
So the answer is (c).