Q:

Consider the recurrence relation and some initial values for the Fibonacci sequence. Fk = Fk βˆ’ 1 + Fk βˆ’ 2 F0 = 1, F1 = 1, F2 = 2, Use the recurrence relation and values for F0, F1, F2, given above to compute F13 and F14.

Accepted Solution

A:
Answer: [tex]F_{13}=377,~~F_{14}=610.[/tex]Step-by-step explanation: Β We are given to consider the following recurrence relation with some initial values for the Fibonacci sequence :[tex]F_k=F_{k-1}+F_{k-2},~~F_0=1,~F_1=1,~F_2=2.[/tex]We are given to use the recurrence relation and given initial values to compute [tex]F_{13}[/tex] and [tex]F_{14}.[/tex]From the given recurrence relation, putting k = 3, 4, . . . , 13, 14, we get[tex]F_3=F_2+F_1=2+1=3,\\\\F_4=F_3+F_2=3+2=5,\\\\F_5=F_4+F_3=5+3=8,\\\\F_6=F_5+F_4=8+5=13,\\\\F_7=F_6+F_5=13+8=21,\\\\F_8=F_7+F_6=21+13=34,\\\\F_9=F_8+F_7=34+21=55,\\\\F_{10}=F_9+F_8=55+34=89,\\\\F_{11}=F_{10}+F_9=89+55=144,\\\\F_{12}=F_{11}+F_{10}=144+89=233,\\\\F_{13}=F_{12}+F_{11}=233+144=377,\\\\F_{14}=F_{13}+F_{12}=377+233=610.[/tex]Thus, [tex]F_{13}=377,~~F_{14}=610.[/tex]