1. **Problem:** Find the least upper bound (supremum) of the set $$A = \left\{ \frac{2}{3^n} + \frac{(-1)^n}{2^{n-1}} \mid n \in \mathbb{N} \right\}$$. 2. **Formula and rules:** The least upper bound is the smallest number that is greater than or equal to every element in the set. 3. **Analyze the terms:** - The first term $$\frac{2}{3^n}$$ is positive and decreases to 0 as $$n \to \infty$$. - The second term $$\frac{(-1)^n}{2^{n-1}}$$ alternates sign and also tends to 0 as $$n \to \infty$$. 4. **Calculate first few terms to identify pattern:** - For $$n=1$$: $$\frac{2}{3^1} + \frac{(-1)^1}{2^{0}} = \frac{2}{3} - 1 = -\frac{1}{3}$$ - For $$n=2$$: $$\frac{2}{3^2} + \frac{(-1)^2}{2^{1}} = \frac{2}{9} + \frac{1}{2} = \frac{4}{18} + \frac{9}{18} = \frac{13}{18}$$ - For $$n=3$$: $$\frac{2}{3^3} + \frac{(-1)^3}{2^{2}} = \frac{2}{27} - \frac{1}{4} = \frac{8}{108} - \frac{27}{108} = -\frac{19}{108}$$ - For $$n=4$$: $$\frac{2}{3^4} + \frac{(-1)^4}{2^{3}} = \frac{2}{81} + \frac{1}{8} = \frac{16}{648} + \frac{81}{648} = \frac{97}{648} \approx 0.1497$$ 5. **Observations:** - The terms alternate above and below zero. - The maximum value among these is at $$n=2$$ with $$\frac{13}{18}$$. - As $$n$$ increases, terms approach 0. 6. **Conclusion:** The least upper bound (supremum) of $$A$$ is $$\boxed{\frac{13}{18}}$$. **Answer:** C. $$\frac{13}{18}$$