1. **Problem statement:** We have two sequences $(U_n)$ defined by recurrence relations and related sequences $(V_n)$ defined in terms of $(U_n)$. We want to show that $(V_n)$ defined by $V_n = \frac{1}{U_n - 3}$ is arithmetic with common difference $d = -\frac{1}{3}$ and find its first term. 2. **Given:** - $U_0 = 1$ - $U_{n+1} = \frac{9}{6 - U_n}$ for all $n \in \mathbb{N}$ - $0 < U_n < 3$ for all $n$ - $U_{n+1} - U_n = \frac{(3 - U_n)^2}{6 - U_n}$ - Define $V_n = \frac{1}{U_n - 3}$ 3. **Goal:** Show $(V_n)$ is arithmetic with common difference $d = -\frac{1}{3}$ and find $V_0$. 4. **Step 1: Express $V_{n+1}$ in terms of $V_n$** Start with $V_n = \frac{1}{U_n - 3}$, so $U_n = 3 + \frac{1}{V_n}$. Using the recurrence for $U_{n+1}$: $$ U_{n+1} = \frac{9}{6 - U_n} = \frac{9}{6 - \left(3 + \frac{1}{V_n}\right)} = \frac{9}{3 - \frac{1}{V_n}} = \frac{9}{\frac{3 V_n - 1}{V_n}} = \frac{9 V_n}{3 V_n - 1}. $$ 5. **Step 2: Write $V_{n+1}$:** $$ V_{n+1} = \frac{1}{U_{n+1} - 3} = \frac{1}{\frac{9 V_n}{3 V_n - 1} - 3} = \frac{1}{\frac{9 V_n - 3(3 V_n - 1)}{3 V_n - 1}} = \frac{1}{\frac{9 V_n - 9 V_n + 3}{3 V_n - 1}} = \frac{1}{\frac{3}{3 V_n - 1}} = \frac{3 V_n - 1}{3}. $$ 6. **Step 3: Simplify $V_{n+1}$:** $$ V_{n+1} = V_n - \frac{1}{3}. $$ This shows $(V_n)$ is an arithmetic sequence with common difference $d = -\frac{1}{3}$. 7. **Step 4: Find $V_0$:** $$ V_0 = \frac{1}{U_0 - 3} = \frac{1}{1 - 3} = \frac{1}{-2} = -\frac{1}{2}. $$ 8. **Summary:** - $(V_n)$ is arithmetic with $V_{n+1} = V_n - \frac{1}{3}$ - First term $V_0 = -\frac{1}{2}$ --- **Additional sequences:** For the second sequence: - $U_0 = 2$ - $U_{n+1} = \frac{2 U_n}{1 - U_n}$ Define $V_n = \left\lfloor \frac{1 + U_n}{U_n} \right\rfloor$ for all $n$. The product $V_0 \times V_3 \times \cdots \times V_{498}$ involves terms of this sequence at indices multiples of 3. --- **Desmos function for first sequence:** $$ y = \frac{9}{6 - x} $$