1. **Problem statement:** Show that the sequence $\{u_n\}$ satisfies the recurrence relation $$u_n = -3u_{n-1} + 4u_{n-2}.$$\n\n2. **Understanding the recurrence relation:** This means each term $u_n$ is defined in terms of the two previous terms $u_{n-1}$ and $u_{n-2}$ by the formula $$u_n = -3u_{n-1} + 4u_{n-2}.$$\n\n3. **To show the sequence is a solution:** We need to verify that for all $n \geq 2$, the given sequence values satisfy the equation above.\n\n4. **Method:** Substitute the known terms $u_{n-1}$ and $u_{n-2}$ into the right side and check if it equals $u_n$.\n\n5. **Example:** Suppose the sequence $\{u_n\}$ is defined or given explicitly (e.g., initial terms or a closed form). Without explicit terms, we can check the characteristic equation associated with the recurrence relation: $$r^2 + 3r - 4 = 0.$$\n\n6. **Solve the characteristic equation:**\n$$r^2 + 3r - 4 = 0$$\nUsing the quadratic formula: $$r = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-4)}}{2 \times 1} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2}.$$\n\n7. **Roots:**\n$$r_1 = \frac{-3 + 5}{2} = 1, \quad r_2 = \frac{-3 - 5}{2} = -4.$$\n\n8. **General solution:** The sequence $\{u_n\}$ can be expressed as $$u_n = A \cdot 1^n + B \cdot (-4)^n = A + B(-4)^n,$$ where $A$ and $B$ are constants determined by initial conditions.\n\n9. **Verification:** Any sequence of this form satisfies the recurrence relation.\n\n**Final answer:** The sequence $\{u_n\}$ defined by $$u_n = A + B(-4)^n$$ is a solution to the recurrence relation $$u_n = -3u_{n-1} + 4u_{n-2}.$$