1. State the problem: Simplify the given expression for $S_n$: $$S_n=3\left[1-\left(\frac{1}{3}\right)^{n}+\frac{1}{2\cdot 3}\right]$$ 2. Simplify the term inside the brackets step-by-step. $$\frac{1}{2\cdot 3}=\frac{1}{6}$$ 3. Substitute and rewrite: $$S_n=3\left[1-\left(\frac{1}{3}\right)^n+\frac{1}{6}\right]$$ 4. Combine the constants inside the brackets: $$1+\frac{1}{6}=\frac{6}{6}+\frac{1}{6}=\frac{7}{6}$$ 5. Thus, $$S_n=3\left[\frac{7}{6}-\left(\frac{1}{3}\right)^n\right]$$ 6. Distribute 3: $$S_n=3\cdot \frac{7}{6} - 3\left(\frac{1}{3}\right)^n = \frac{21}{6} - 3 \left(\frac{1}{3}\right)^n = \frac{7}{2} - 3 \left(\frac{1}{3}\right)^n$$ 7. Simplify the term \(3 \left(\frac{1}{3}\right)^n\): Since $3 = 3^1$, $$3 \left(\frac{1}{3}\right)^n = 3^1 \cdot 3^{-n} = 3^{1 - n}$$ 8. Final simplified form: $$S_n = \frac{7}{2} - 3^{1 - n}$$