1. **State the problem:** We have a sequence defined by the recurrence relation $$u_{n+3} = u_n + 2$$ with initial values $$u_1 = 0$$, $$u_2 = 1$$, and $$u_3 = 5$$. We need to find the index $$k$$ such that $$u_k = 200$$. 2. **Understand the recurrence:** The term $$u_{n+3}$$ depends on $$u_n$$ plus 2. This means every term three places ahead increases by 2 compared to the term at position $$n$$. 3. **Find the pattern:** Let's write out the first few terms: - $$u_1 = 0$$ - $$u_2 = 1$$ - $$u_3 = 5$$ - $$u_4 = u_1 + 2 = 0 + 2 = 2$$ - $$u_5 = u_2 + 2 = 1 + 2 = 3$$ - $$u_6 = u_3 + 2 = 5 + 2 = 7$$ - $$u_7 = u_4 + 2 = 2 + 2 = 4$$ - $$u_8 = u_5 + 2 = 3 + 2 = 5$$ - $$u_9 = u_6 + 2 = 7 + 2 = 9$$ 4. **Observe subsequences:** The sequence splits into three subsequences based on $$n mod 3$$: - For $$n = 3m + 1$$: $$u_1 = 0$$, $$u_4 = 2$$, $$u_7 = 4$$, ... - For $$n = 3m + 2$$: $$u_2 = 1$$, $$u_5 = 3$$, $$u_8 = 5$$, ... - For $$n = 3m$$: $$u_3 = 5$$, $$u_6 = 7$$, $$u_9 = 9$$, ... Each subsequence is arithmetic with common difference 2. 5. **General formulas:** - If $$n = 3m + 1$$, then $$u_n = 0 + 2m = 2m$$ - If $$n = 3m + 2$$, then $$u_n = 1 + 2m$$ - If $$n = 3m$$, then $$u_n = 5 + 2(m-1) = 2m + 3$$ 6. **Find $$k$$ such that $$u_k = 200$$:** Check each case: - Case 1: $$u_n = 2m = 200 \Rightarrow m = 100$$, so $$n = 3m + 1 = 3 \times 100 + 1 = 301$$ - Case 2: $$u_n = 1 + 2m = 200 \Rightarrow 2m = 199 \Rightarrow m = 99.5$$ (not integer, discard) - Case 3: $$u_n = 2m + 3 = 200 \Rightarrow 2m = 197 \Rightarrow m = 98.5$$ (not integer, discard) 7. **Conclusion:** The only valid solution is $$k = 301$$. **Final answer:** $$\boxed{301}$$