Answer:
a) f(2) = 0; f(3) = f(4) = -1; f(5) = 0
b) f(2)=f(3)=f(4)=f(5) = 1
c) f(2) = 2; f(3) = 5; f(4) = 27; f(5) = 734
d) f(2)=f(3)=f(4)=f(5) = 1
Step-by-step explanation:
In general we have f(0) = f(1) = 1
a) We have
f(n+1) = f(n)-f(n-1)
Thus,
f(2) = f(1)-f(0) = 1-1 = 0
f(3) = f(2)-f(1) = 0-1 = -1
f(4) = f(3)-f(2) = -1-0 = -1
f(5) = f(4)-f(3) = -1-(-1) = 0
b) Here instead we have
f(n+1) = f(n)-f(n-1)
If f(n) = f(n-1) = 1, then f(n+1) should be 1, thus
f(2) = 1*1 = 1
f(3) = 1*1 = 1
f(4) = 1*1 = 1
f(5) = 1*1 = 1.
We conclude that f(2) = f(3) = f(4) = f(5) = 1
c) In this one we have
f(n+1) = f(n)²+f(n-1)
f(2) = f(1)²+f(0) = 1²+1 = 2
f(3) = f(2)²+f(1) = 2²+1 = 5
f(4) = f(3)²+f(2) = 5²+2 = 27
f(5) = f(4)²+f(3) = 27²+5 = 734
d) Now
f(n+1) = f(n)/f(n-1)
f(2) = f(1)/f(0) = 1/1 = 1
f(3) = f(2)/f(1) = 1/1 = 1
f(4) = f(3)/f(2) = 1/1 = 1
f(5) = f(4)/f(3) = 1/1 = 1.
Thus, f(2) = f(3) = f(4) = f(5) = 1.
