1. **Problem Statement:** Prove by induction that the sum of the series $$1 \cdot 3^0 + 2 \cdot 3^1 + 3 \cdot 3^2 + \cdots + n \cdot 3^{n-1} = \frac{n \cdot 3^{n+1} - (n + 1)3^n + 1}{4}$$ 2. **Base Case (n=1):** Calculate the left side: $$1 \cdot 3^0 = 1$$ Calculate the right side: $$\frac{1 \cdot 3^{2} - 2 \cdot 3^{1} + 1}{4} = \frac{1 \cdot 9 - 2 \cdot 3 + 1}{4} = \frac{9 - 6 + 1}{4} = \frac{4}{4} = 1$$ Since both sides equal 1, the base case holds. 3. **Inductive Hypothesis:** Assume the formula holds for some integer $k \geq 1$: $$\sum_{i=1}^k i \cdot 3^{i-1} = \frac{k \cdot 3^{k+1} - (k + 1)3^k + 1}{4}$$ 4. **Inductive Step:** We need to prove it holds for $k+1$: $$\sum_{i=1}^{k+1} i \cdot 3^{i-1} = \sum_{i=1}^k i \cdot 3^{i-1} + (k+1) \cdot 3^k$$ Using the inductive hypothesis: $$= \frac{k \cdot 3^{k+1} - (k + 1)3^k + 1}{4} + (k+1) \cdot 3^k$$ Rewrite $(k+1) \cdot 3^k$ as $\frac{4(k+1)3^k}{4}$ to combine terms: $$= \frac{k \cdot 3^{k+1} - (k + 1)3^k + 1 + 4(k+1)3^k}{4}$$ Simplify the numerator: $$k \cdot 3^{k+1} - (k + 1)3^k + 1 + 4(k+1)3^k = k \cdot 3^{k+1} + 3(k+1)3^k + 1$$ Note that $3(k+1)3^k = (k+1)3^{k+1}$, so: $$= k \cdot 3^{k+1} + (k+1)3^{k+1} + 1 = (k + k + 1)3^{k+1} + 1 = (2k + 1)3^{k+1} + 1$$ 5. **Rewrite the right side of the formula for $k+1$:** $$\frac{(k+1) \cdot 3^{k+2} - (k + 2)3^{k+1} + 1}{4}$$ Factor $3^{k+1}$: $$= \frac{3^{k+1} \left((k+1)3 - (k+2)\right) + 1}{4} = \frac{3^{k+1} (3k + 3 - k - 2) + 1}{4} = \frac{3^{k+1} (2k + 1) + 1}{4}$$ 6. **Conclusion:** The expression from the inductive step matches the formula for $k+1$, so by mathematical induction, the formula is true for all positive integers $n$. **Final answer:** $$\sum_{i=1}^n i \cdot 3^{i-1} = \frac{n \cdot 3^{n+1} - (n + 1)3^n + 1}{4}$$