Find f( 2 ) , f( 3 ) , f( 4 ) , and f( 5 ) if f is defined recur-sively by f( 0 ) = f( 1 ) = 1 and for n = 1 , 2 a) f(n + 1 ) = f(n) − f(n − 1 ) . b) f(n + 1 ) = f(n)f(n − 1 ) . c) f(n + 1 ) = f(n) 2 + f(n − 1 ) 3 . d) f(n + 1 ) = f(n)/f(n − 1 ) . slader

(a) f(2) = 0, f(3) = -1, f(4) = -1, f(5) = 0

(b) f(2) = 1, f(3) = 1, f(4) = 1, f(5) = 1

(c) f(2) = 5, f(3) = 13, f(4) = 41, f(5) = 121

(d) f(2) = 1, f(3) = 1, f(4) = 1, f(5) = 1

Step-by-step explanation:

(a)

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)

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

(c)

f(2) = 2f(1) + 3f(0) = 2 + 3 = 5

f(3) = 2f(2) + 3f(1) = 10 + 3 = 13

f(4) = 2f(3) + 3f(2) = 26 + 15 = 41

f(5) = 2f(4) + 3f(3) = 82 + 39 = 121

(d)

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

i am thinking c. sorry if im wrong .

step-by-step explanation: