DSC01318 (2)

4.b Algorytm rekurencyjny obliczania F(n)

0 dla n = 0    Obliczmy kilka pierwszych

F(n) = <    1 dla n = l    wyrazów ciągu.

[F(n-1)+ F(n 2) dla n >l[

») = u

F(1) = 1

F(2) = F(1) + F(0) = 1 + 0 = 1 F(3) = F(2) + F(1) = (F(1) + F(0)) + F(1) = 1 + 0 +1 = 2 I _F(4) = F(3) + F(2) = (F(2) + F(1)) + (F(1) + F(0)) = „

IL (F(1) + F(0)+F(1)) + (F(1) + F(0)) = 1 + 0+1+1 + 0 = 3 i If(5) = F(4) + F(3) = (F(3) + F(2)) + (F(2) + F(1)) =

= (F(2) + F(1) + F(1) + F(0)) + (F(1) + F(0) + F(1)) =........

i F(6) = F(5) + F(41 = (F(4) + F(3)) + (F(3) + F(2))

ILlEiaLhHas Bil+F<m+(F(2i+Fm+F(1)+F(0»