(a) Compute the values of \( u_{1} \) and \( u_{2} \) when the difference equations and initial values are as given below. (i) \( u_{n+1}=4 u_{n}\left(u_{n}-5\right), u_{0}=7 \) (ii) \( u_{n+1}=\frac{u_{n}}{3}-\frac{1}{2 u_{n}}, u_{0}=1 \).

Let's compute the values of \( u_1 \) and \( u_2 \) for each of the given difference equations. ### (i) \( u_{n+1} = 4 u_n (u_n - 5) \) with \( u_0 = 7 \) 1. **Compute \( u_1 \):** \[ u_1 = 4 u_0 (u_0 - 5) = 4 \times 7 \times (7 - 5) = 4 \times 7 \times 2 = 56 \] 2. **Compute \( u_2 \):** \[ u_2 = 4 u_1 (u_1 - 5) = 4 \times 56 \times (56 - 5) = 4 \times 56 \times 51 \] \[ 56 \times 51 = 2856 \quad \text{(since } 56 \times 50 = 2800 \text{ and } 56 \times 1 = 56\text{)} \] \[ u_2 = 4 \times 2856 = 11424 \] **Results for (i):** - \( u_1 = 56 \) - \( u_2 = 11424 \) --- ### (ii) \( u_{n+1} = \frac{u_n}{3} - \frac{1}{2 u_n} \) with \( u_0 = 1 \) 1. **Compute \( u_1 \):** \[ u_1 = \frac{u_0}{3} - \frac{1}{2 u_0} = \frac{1}{3} - \frac{1}{2 \times 1} = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6} \] 2. **Compute \( u_2 \):** \[ u_2 = \frac{u_1}{3} - \frac{1}{2 u_1} = \frac{-\frac{1}{6}}{3} - \frac{1}{2 \times \left(-\frac{1}{6}\right)} = -\frac{1}{18} + 3 = \frac{53}{18} \] \[ \text{(Since } 3 = \frac{54}{18} \text{ and } \frac{54}{18} - \frac{1}{18} = \frac{53}{18}\text{)} \] **Results for (ii):** - \( u_1 = -\frac{1}{6} \) - \( u_2 = \frac{53}{18} \) --- **Final Answer:** Problem i Answer:  \( u_{1} = 56 \) and \( u_{2} = 11424 \). Problem ii Answer:  \( u_{1} = -\dfrac{1}{6} \) and \( u_{2} = \dfrac{53}{18} \).