1. **Statement of the problem:** We have two sequences defined as follows: $$U_0 = -1$$ $$U_{n+1} = \frac{3}{2} U_n - 1$$ and $$V_n = U_n - 2$$ We need to: (1) Calculate $4, 4^2, 4^3, 4^4$. (2) Calculate $\frac{V_{n+1}}{V_n}$ and deduce that $(V_n)$ is a geometric sequence with ratio $\frac{3}{2}$ and find its first term. (3) Express $V_n$ in terms of $n$. (4) Deduce $U_n$ in terms of $n$. (5) Calculate the limits $\lim_{n \to +\infty} U_n$ and $\lim_{n \to +\infty} V_n$. 2. **Step 1: Calculate powers of 4** $$4 = 4$$ $$4^2 = 16$$ $$4^3 = 64$$ $$4^4 = 256$$ 3. **Step 2: Calculate $\frac{V_{n+1}}{V_n}$** Recall $V_n = U_n - 2$. From the recurrence: $$U_{n+1} = \frac{3}{2} U_n - 1$$ Subtract 2 from both sides: $$U_{n+1} - 2 = \frac{3}{2} U_n - 1 - 2 = \frac{3}{2} U_n - 3$$ But $V_{n+1} = U_{n+1} - 2$, so: $$V_{n+1} = \frac{3}{2} U_n - 3$$ Rewrite $U_n$ as $V_n + 2$: $$V_{n+1} = \frac{3}{2} (V_n + 2) - 3 = \frac{3}{2} V_n + 3 - 3 = \frac{3}{2} V_n$$ Therefore: $$\frac{V_{n+1}}{V_n} = \frac{3}{2}$$ This shows $(V_n)$ is a geometric sequence with ratio $q = \frac{3}{2}$. 4. **Step 3: Find the first term $V_0$** $$V_0 = U_0 - 2 = -1 - 2 = -3$$ 5. **Step 4: Express $V_n$ in terms of $n$** Since $(V_n)$ is geometric: $$V_n = V_0 \times q^n = -3 \times \left(\frac{3}{2}\right)^n$$ 6. **Step 5: Express $U_n$ in terms of $n$** Recall: $$U_n = V_n + 2 = -3 \left(\frac{3}{2}\right)^n + 2$$ 7. **Step 6: Calculate limits** - Since $\left|\frac{3}{2}\right| > 1$, $\left(\frac{3}{2}\right)^n \to +\infty$ as $n \to +\infty$. - Therefore: $$\lim_{n \to +\infty} V_n = \lim_{n \to +\infty} -3 \left(\frac{3}{2}\right)^n = -\infty$$ $$\lim_{n \to +\infty} U_n = \lim_{n \to +\infty} \left(-3 \left(\frac{3}{2}\right)^n + 2\right) = -\infty$$ --- **Exercise 2: Calculate limits** 1. $$\lim_{n \to +\infty} (n^2 - 5n^3 + 4)$$ Dominated by $-5n^3$, so limit is $-\infty$. 2. $$\lim_{n \to +\infty} \frac{7n^2 + 3n + 1}{n^5 + 3}$$ Degree numerator 2, denominator 5, so limit is 0. 3. $$\lim_{n \to +\infty} \frac{6n^2 + 8n + 7}{n^4 + 3}$$ Degree numerator 2, denominator 4, so limit is 0. 4. $$\lim_{n \to +\infty} \frac{7n^4 + 2n - 1}{n^4 + 9}$$ Divide numerator and denominator by $n^4$: $$\lim_{n \to +\infty} \frac{7 + \frac{2}{n^3} - \frac{1}{n^4}}{1 + \frac{9}{n^4}} = 7$$ 5. $$\lim_{n \to +\infty} \left(\frac{5}{n} - 1\right) \left(\frac{1}{\sqrt{n}} + 2\right)$$ As $n \to +\infty$, $\frac{5}{n} \to 0$ and $\frac{1}{\sqrt{n}} \to 0$, so: $$(-1)(2) = -2$$ 6. $$\lim_{n \to +\infty} n^n - 3^n$$ Since $n^n$ grows faster than $3^n$, limit is $+\infty$. **Final answers:** - Exercise 1: $$V_n = -3 \left(\frac{3}{2}\right)^n$$ $$U_n = -3 \left(\frac{3}{2}\right)^n + 2$$ $$\lim_{n \to +\infty} V_n = -\infty$$ $$\lim_{n \to +\infty} U_n = -\infty$$ - Exercise 2: 1) $-\infty$ 2) $0$ 3) $0$ 4) $7$ 5) $-2$ 6) $+\infty$