1. **Problem Statement:** Verify that the sequence $$u_n = \frac{14}{3} \left(\frac{1}{2}\right)^n + l$$ satisfies both the initial condition and the recurrence relation given in the question for a certain value of $l$. 2. **Given:** - The sequence formula: $$u_n = \frac{14}{3} \left(\frac{1}{2}\right)^n + l$$ - We need to find $l$ such that the sequence satisfies the initial condition and the recurrence relation. 3. **Step 1: Verify initial condition** Assuming the initial condition is $u_0 = 7$ (from the context of the problem), substitute $n=0$: $$u_0 = \frac{14}{3} \left(\frac{1}{2}\right)^0 + l = \frac{14}{3} + l$$ Set equal to 7: $$\frac{14}{3} + l = 7$$ Solve for $l$: $$l = 7 - \frac{14}{3} = \frac{21}{3} - \frac{14}{3} = \frac{7}{3}$$ 4. **Step 2: Verify the recurrence relation** Assuming the recurrence relation is: $$u_{n+1} = \frac{1}{2} u_n + c$$ for some constant $c$. Substitute $u_n$ and $u_{n+1}$: $$u_{n+1} = \frac{14}{3} \left(\frac{1}{2}\right)^{n+1} + l$$ $$u_n = \frac{14}{3} \left(\frac{1}{2}\right)^n + l$$ Check if: $$\frac{14}{3} \left(\frac{1}{2}\right)^{n+1} + l = \frac{1}{2} \left( \frac{14}{3} \left(\frac{1}{2}\right)^n + l \right) + c$$ Simplify right side: $$= \frac{1}{2} \cdot \frac{14}{3} \left(\frac{1}{2}\right)^n + \frac{1}{2} l + c = \frac{14}{3} \left(\frac{1}{2}\right)^{n+1} + \frac{1}{2} l + c$$ Equate left and right sides: $$\frac{14}{3} \left(\frac{1}{2}\right)^{n+1} + l = \frac{14}{3} \left(\frac{1}{2}\right)^{n+1} + \frac{1}{2} l + c$$ Cancel common terms: $$l = \frac{1}{2} l + c$$ Solve for $c$: $$c = l - \frac{1}{2} l = \frac{1}{2} l = \frac{1}{2} \times \frac{7}{3} = \frac{7}{6}$$ 5. **Conclusion:** - The value of $l$ is $$\frac{7}{3}$$. - The recurrence relation is: $$u_{n+1} = \frac{1}{2} u_n + \frac{7}{6}$$ - The sequence $$u_n = \frac{14}{3} \left(\frac{1}{2}\right)^n + \frac{7}{3}$$ satisfies both the initial condition and the recurrence relation.