Define a sequence recursively by \(F_{0}=0,~F_{1}=1\) and \(F_{n}\) be the remainder when \(F_{n-1}+F_{n-2}\) is divided by \(3\) for all \(n\geq 2\) Thus the sequence starts \(0,1,1,2,0,2,\ldots \) What is \(F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?\)
Please help me! I really would appreciate it :)
+128474 Let's see if any pattern "falls out"
[0 + 1 ] / 3 = 1 = F(2)
[ 1 + 1] / 3 = 2 = F(3)
[ 1 + 2 ] / 3 = 0 = F(4)
[ 2 + 0 ] / 3 = 2 = F(5)
[ 0 + 2 ] / 3 = 2 = F(6)
[ 2 + 2] / 3 = 1 = F(7)
[ 2+ 1 ] / 3 = 0 = F(8)
[ 1 + 0 ] / 3 = 1 = F(9)
[ 0 + 1] / 3 = 1 = F(10)
[ 1 + 1] / 3 = 2 = F(11)
[ 1 + 2 ] / 3 = 0 = F(12)
[ 2 + 0 ] / 3 = 2 = F(13)
[ 0 + 2] / 3 = 2 = F(14)
[2 + 2 ] / 3 = 1 = F(15)
[ 2 + 1] / 3 = 0 = F(16)
[ 1 + 0] / 3 = 1 = F(17)
Note that the pattern has a cycle of 8
And the cycle starts at the indexes 2 , 10 , 18, 26 ...etc.
So the pattern begins repeats for every [-6 + 8n ] th index number
And note that
-6 + 8(253) = 2018
So this means that
F(2018) begins the 253nd cycle
So this must mean that
F(2017) = 1 [ the last remainder in the 252nd cycle ]
F(2018) = 1
F(2019) = 2
F(2020) = 0
F(2021) = 2
F(2022) = 2
F(2023) = 1
F(2024) = 0
And the sum of these remainders = 9
