1. **State the problem:** Prove by mathematical induction that for all positive integers $n$, $$1 + 4 + 4^2 + \cdots + 4^n = \frac{1}{3}(4^{n+1} - 1)$$ 2. **Base case ($n=1$):** Left side: $1 + 4 = 5$ Right side: $\frac{1}{3}(4^{1+1} - 1) = \frac{1}{3}(4^2 - 1) = \frac{1}{3}(16 - 1) = \frac{15}{3} = 5$ Since both sides equal 5, the base case holds. 3. **Inductive hypothesis:** Assume the formula holds for some positive integer $k$, i.e., $$1 + 4 + 4^2 + \cdots + 4^k = \frac{1}{3}(4^{k+1} - 1)$$ 4. **Inductive step:** Prove the formula holds for $k+1$: Consider the sum up to $k+1$: $$1 + 4 + 4^2 + \cdots + 4^k + 4^{k+1}$$ Using the inductive hypothesis, this equals $$\frac{1}{3}(4^{k+1} - 1) + 4^{k+1}$$ Rewrite $4^{k+1}$ as $\frac{3}{3}4^{k+1}$ to combine terms: $$\frac{1}{3}(4^{k+1} - 1) + \frac{3}{3}4^{k+1} = \frac{1}{3}(4^{k+1} - 1) + \frac{3}{3}4^{k+1}$$ Combine the fractions: $$= \frac{1}{3}(4^{k+1} - 1) + \frac{3}{3}4^{k+1} = \frac{1}{3}(4^{k+1} - 1) + \frac{3}{3}4^{k+1} = \frac{1}{3}(4^{k+1} - 1 + 3 \times 4^{k+1})$$ Simplify inside the parentheses: $$4^{k+1} - 1 + 3 \times 4^{k+1} = 4^{k+1} - 1 + 3 \cdot 4^{k+1} = 4^{k+1} + 3 \cdot 4^{k+1} - 1 = 4 \cdot 4^{k+1} - 1 = 4^{k+2} - 1$$ So the sum is: $$\frac{1}{3}(4^{k+2} - 1)$$ This matches the formula for $n = k+1$. 5. **Conclusion:** By mathematical induction, the formula $$1 + 4 + 4^2 + \cdots + 4^n = \frac{1}{3}(4^{n+1} - 1)$$ holds for all positive integers $n$.