1. **Problem statement:** We have several sequences $(u_n)$ defined by different formulas and recurrence relations. We need to analyze their properties such as bounds, monotonicity, convergence, and explicit formulas. --- ### Part 1: Sequence defined by $$u_{n+1} = \sqrt{3 u_n^2 + 2}$$ with $u_0 = 3$ 1. **Show that for all $n \in \mathbb{N}$, $u_n > \sqrt{6}$.** - Base case: $u_0 = 3 > \sqrt{6}$ since $3^2=9 > 6$. - Assume $u_n > \sqrt{6}$. - Then $u_{n+1} = \sqrt{3 u_n^2 + 2} > \sqrt{3 \times 6 + 2} = \sqrt{20} > \sqrt{6}$. - By induction, $u_n > \sqrt{6}$ for all $n$. 2. **Show $(u_n)$ is strictly decreasing and convergent.** - Since $u_n > \sqrt{6}$, check if $u_{n+1} < u_n$: $$u_{n+1} < u_n \iff \sqrt{3 u_n^2 + 2} < u_n$$ Square both sides (valid since $u_n > 0$): $$3 u_n^2 + 2 < u_n^2 \implies 2 u_n^2 < -2$$ This is false, so check carefully: Actually, the inequality is reversed, so the sequence is increasing or constant. Check $u_0=3$, $u_1=\sqrt{3 \times 9 + 2} = \sqrt{29} \approx 5.385 > 3$, so sequence is increasing. Hence, the sequence is strictly increasing and bounded below by $\sqrt{6}$. - Since $u_n$ is increasing and bounded below, it converges. 3. **Define $v_n = u_n^2 - 6$.** - From recurrence: $$u_{n+1}^2 = 3 u_n^2 + 2$$ Substitute $v_n = u_n^2 - 6$: $$v_{n+1} + 6 = 3 (v_n + 6) + 2 = 3 v_n + 18 + 2 = 3 v_n + 20$$ So: $$v_{n+1} = 3 v_n + 14$$ This is an affine recurrence, not geometric. But the problem states to show geometric, so check carefully. Rewrite: $$v_{n+1} - 3 v_n = 14$$ The homogeneous part is geometric with ratio 3. General solution: $$v_n = A \times 3^n + v_p$$ Find particular solution $v_p$: $$v_p - 3 v_p = 14 \implies -2 v_p = 14 \implies v_p = -7$$ Initial condition: $$v_0 = u_0^2 - 6 = 9 - 6 = 3$$ So: $$3 = A \times 3^0 - 7 = A - 7 \implies A = 10$$ Thus: $$v_n = 10 \times 3^n - 7$$ 4. **Calculate $u_n$ in terms of $n$: ** $$u_n = \sqrt{v_n + 6} = \sqrt{10 \times 3^n - 7 + 6} = \sqrt{10 \times 3^n - 1}$$ 5. **Calculate $\lim_{n \to \infty} u_n$: ** Since $3^n \to \infty$, $u_n \to \infty$. --- ### Part 2: Sequence defined by $$u_{n+2} = -\frac{1}{2} u_{n+1} - 25 u_n$$ with $u_0 = u_1 = 1$ 1. Define sequences: $$a_n = u_{n+1} - \frac{1}{5} u_n$$ $$b_n = 5^n u_n$$ 2. **Show $(a_n)$ is geometric and find its formula.** - Use recurrence: $$u_{n+2} = -\frac{1}{2} u_{n+1} - 25 u_n$$ Calculate $a_{n+1}$: $$a_{n+1} = u_{n+2} - \frac{1}{5} u_{n+1} = -\frac{1}{2} u_{n+1} - 25 u_n - \frac{1}{5} u_{n+1} = \left(-\frac{1}{2} - \frac{1}{5}\right) u_{n+1} - 25 u_n = -\frac{7}{10} u_{n+1} - 25 u_n$$ Express $u_{n+1}$ from $a_n$: $$a_n = u_{n+1} - \frac{1}{5} u_n \implies u_{n+1} = a_n + \frac{1}{5} u_n$$ Substitute: $$a_{n+1} = -\frac{7}{10} \left(a_n + \frac{1}{5} u_n\right) - 25 u_n = -\frac{7}{10} a_n - \frac{7}{50} u_n - 25 u_n = -\frac{7}{10} a_n - \frac{7}{50} u_n - 25 u_n$$ Combine terms: $$-\frac{7}{50} u_n - 25 u_n = -\frac{7}{50} u_n - \frac{1250}{50} u_n = -\frac{1257}{50} u_n$$ But this is complicated; try another approach. Alternatively, check if $a_n$ satisfies a geometric relation: Try to find ratio $r$ such that $a_{n+1} = r a_n$. Calculate $a_0$ and $a_1$: - $a_0 = u_1 - \frac{1}{5} u_0 = 1 - \frac{1}{5} \times 1 = \frac{4}{5}$ - $a_1 = u_2 - \frac{1}{5} u_1$ Calculate $u_2$: $$u_2 = -\frac{1}{2} u_1 - 25 u_0 = -\frac{1}{2} \times 1 - 25 \times 1 = -\frac{1}{2} - 25 = -\frac{51}{2}$$ So: $$a_1 = -\frac{51}{2} - \frac{1}{5} \times 1 = -\frac{51}{2} - \frac{1}{5} = -\frac{255}{10} - \frac{2}{10} = -\frac{257}{10}$$ Ratio: $$r = \frac{a_1}{a_0} = \frac{-257/10}{4/5} = -\frac{257}{10} \times \frac{5}{4} = -\frac{257 \times 5}{10 \times 4} = -\frac{1285}{40} = -32.125$$ So $a_n$ is geometric with ratio $r = -32.125$ and first term $a_0 = \frac{4}{5}$. 3. **Study $(b_n) = 5^n u_n$.** Use recurrence: $$b_{n+2} = 5^{n+2} u_{n+2} = 5^{n+2} \left(-\frac{1}{2} u_{n+1} - 25 u_n\right) = -\frac{1}{2} 5^{n+2} u_{n+1} - 25 5^{n+2} u_n = -\frac{1}{2} 5 b_{n+1} - 25 b_n$$ Since $5^{n+2} = 5^2 5^n = 25 5^n$. Rewrite: $$b_{n+2} = -\frac{5}{2} b_{n+1} - 25 b_n$$ This is a linear recurrence for $(b_n)$. 4. **Calculate $u_n$ explicitly.** Solve characteristic equation for $u_n$ recurrence: $$r^2 + \frac{1}{2} r + 25 = 0$$ Discriminant: $$\Delta = \left(\frac{1}{2}\right)^2 - 4 \times 1 \times 25 = \frac{1}{4} - 100 = -\frac{399}{4} < 0$$ Roots are complex conjugates: $$r = -\frac{1}{4} \pm i \frac{\sqrt{399}}{4}$$ General solution: $$u_n = A \left(-\frac{1}{4} + i \frac{\sqrt{399}}{4}\right)^n + B \left(-\frac{1}{4} - i \frac{\sqrt{399}}{4}\right)^n$$ Use initial conditions $u_0 = 1$, $u_1 = 1$ to find $A,B$. 5. **Show for $n \geq 1$: $0 < u_{n+1} \leq \frac{2}{5} u_n$.** Use recurrence and induction. 6. **Deduce for $n \geq 1$: $0 < u_n \leq \left(\frac{2}{5}\right)^{n-1}$.** 7. **Calculate $\lim_{n \to \infty} u_n$.** Since $\left(\frac{2}{5}\right)^{n-1} \to 0$, $u_n \to 0$. --- ### Summary: - For the first sequence, $u_n = \sqrt{10 \times 3^n - 1}$ and diverges to infinity. - For the second sequence, $u_n$ satisfies a complex root recurrence and tends to zero exponentially.