1. **Statement of the problem:** We have a sequence $(u_n)$ defined by $u_0 = -1$ and the recurrence relation $$u_{n+1} = \frac{3}{2} u_n - 1.$$ We define another sequence $(v_n)$ by $$v_n = u_n - 2.$$ We need to calculate $u_1$, $u_2$, $u_0$, $v_0$, $v_1$. 2. **Calculate $u_1$ and $u_2$:** $$u_1 = \frac{3}{2} u_0 - 1 = \frac{3}{2} \times (-1) - 1 = -\frac{3}{2} - 1 = -\frac{5}{2}.$$ $$u_2 = \frac{3}{2} u_1 - 1 = \frac{3}{2} \times \left(-\frac{5}{2}\right) - 1 = -\frac{15}{4} - 1 = -\frac{19}{4}.$$ 3. **Calculate $v_0$ and $v_1$:** $$v_0 = u_0 - 2 = -1 - 2 = -3.$$ $$v_1 = u_1 - 2 = -\frac{5}{2} - 2 = -\frac{5}{2} - \frac{4}{2} = -\frac{9}{2}.$$ 4. **Calculate the ratio $\frac{v_{n+1}}{v_n}$:** From the definition, $$v_{n+1} = u_{n+1} - 2 = \frac{3}{2} u_n - 1 - 2 = \frac{3}{2} u_n - 3.$$ But since $v_n = u_n - 2$, then $u_n = v_n + 2$, so $$v_{n+1} = \frac{3}{2} (v_n + 2) - 3 = \frac{3}{2} v_n + 3 - 3 = \frac{3}{2} v_n.$$ Therefore, $$\frac{v_{n+1}}{v_n} = \frac{3}{2}.$$ 5. **Conclusion:** The sequence $(v_n)$ is geometric with ratio $\frac{3}{2}$ and first term $v_0 = -3$. 6. **Express $v_n$ in terms of $n$:** Since $(v_n)$ is geometric, $$v_n = v_0 \left(\frac{3}{2}\right)^n = -3 \left(\frac{3}{2}\right)^n.$$ 7. **Express $u_n$ in terms of $n$:** Recall $u_n = v_n + 2$, so $$u_n = -3 \left(\frac{3}{2}\right)^n + 2.$$ 8. **Calculate limits:** - Since $\frac{3}{2} > 1$, $\lim_{n \to +\infty} \left(\frac{3}{2}\right)^n = +\infty$. - Therefore, $$\lim_{n \to +\infty} u_n = \lim_{n \to +\infty} \left(-3 \left(\frac{3}{2}\right)^n + 2\right) = -3 \times +\infty + 2 = -\infty.$$ - And $$\lim_{n \to +\infty} v_n = \lim_{n \to +\infty} -3 \left(\frac{3}{2}\right)^n = -\infty.$$