1. **Problem Statement:** Prove by mathematical induction that $$3 + 3^2 + 3^3 + \cdots + 3^n = 0.5(3^{n+1} - 3)$$ for all positive integers $n$. 2. **Base Case:** Check for $n=1$. L.H.S. = $3$ R.H.S. = $0.5(3^{1+1} - 3) = 0.5(3^2 - 3) = 0.5(9 - 3) = 0.5 \times 6 = 3$ Since L.H.S. = R.H.S., the base case holds. 3. **Inductive Hypothesis:** Assume the formula holds for some positive integer $k$, i.e., $$3 + 3^2 + 3^3 + \cdots + 3^k = 0.5(3^{k+1} - 3)$$ 4. **Inductive Step:** Prove the formula holds for $k+1$. Consider the sum up to $k+1$: $$3 + 3^2 + 3^3 + \cdots + 3^k + 3^{k+1}$$ Using the inductive hypothesis, this equals: $$0.5(3^{k+1} - 3) + 3^{k+1}$$ Simplify the right side: $$0.5(3^{k+1} - 3) + 3^{k+1} = 0.5 \times 3^{k+1} - 1.5 + 3^{k+1} = 1.5 \times 3^{k+1} - 1.5$$ Rewrite $1.5$ as $\frac{3}{2}$: $$= \frac{3}{2} 3^{k+1} - \frac{3}{2} = \frac{3}{2}(3^{k+1} - 1)$$ Note that: $$3^{k+2} = 3 \times 3^{k+1}$$ So, $$0.5(3^{k+2} - 3) = 0.5(3 \times 3^{k+1} - 3) = 0.5 \times 3 (3^{k+1} - 1) = \frac{3}{2}(3^{k+1} - 1)$$ This matches the expression obtained for the sum up to $k+1$. 5. **Conclusion:** Since the base case holds and the inductive step is true, by mathematical induction, the formula $$3 + 3^2 + 3^3 + \cdots + 3^n = 0.5(3^{n+1} - 3)$$ is true for all positive integers $n$.