Recursion Story Begins!

${ F }_{ k }=\left\{ 1,1,2,0,2,2,1,0,1,1,........ \right\}$

we can see that

${ F }_{ k }=\begin{cases} 1 & & k=8n \\ 1 & & k=8n+1 \\ 2 & & k=8n+2 \\ 0 & & k=8n+3 \\ 2 & & k=8n+4 \\ 2 & & k=8n+5 \\ 1 & & k=8n+6 \\ 0 & & k=8n+7 \end{cases}$

so $\sum _{ k=0 }^{ 8 }{ { F }_{ k } } =9\Rightarrow \sum _{ k=0 }^{ 8n }{ { F }_{ k } } =9n$

$\sum _{ k=2017 }^{ 2024 }{ { F }_{ k } } =\sum _{ k=0 }^{ 2024 }{ { F }_{ k } } -\sum _{ k=0 }^{ 2016 }{ { F }_{ k } }$

$=\sum _{ k=0 }^{ 8\cdot 253 }{ { F }_{ k } } -\sum _{ k=0 }^{ 8\cdot 252 }{ { F }_{ k } } =9\times 253-9\times 252=9$

This sequence repeats itself after 8 terms: 1, 1, 2, 0, 2, 2, 1, 0, and so on. To find $F_k$ where k=2017, we divide 2017 by 8, which yields a remainder of 1. The first term is 1. 2024 divided by 8 yields a remainder of 0 (which is basically a remainder of 8). The 8th term is 0. So the answer is the sum of all the terms in one non-repeating sequence, which is 1 + 1 + 2 + 0 + 2 + 2 + 1 + 0 = $\boxed{9}$ .