1. **Problem Statement:** We want to prove the formula for the sum of the first $n$ squares: $$1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$ for all positive integers $n$. 2. **Part (a): What is the statement $P(1)$?** The statement $P(1)$ is: $$1^2 = \frac{1(1+1)(2\cdot1+1)}{6}$$ which simplifies to: $$1 = \frac{1 \times 2 \times 3}{6}$$ 3. **Part (b): Show that $P(1)$ is true (basis step).** Calculate the right side: $$\frac{1 \times 2 \times 3}{6} = \frac{6}{6} = 1$$ The left side is $1^2 = 1$, so both sides are equal. Thus, $P(1)$ is true. 4. **Part (c): What is the inductive hypothesis?** Assume $P(k)$ is true for some positive integer $k$, i.e., $$1^2 + 2^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6}$$ 5. **Part (d): What do you need to prove in the inductive step?** We need to prove $P(k+1)$ is true, that is: $$1^2 + 2^2 + \cdots + k^2 + (k+1)^2 = \frac{(k+1)(k+2)(2(k+1)+1)}{6}$$ 6. **Part (e): Complete the inductive step using the inductive hypothesis.** Start with the left side of $P(k+1)$: $$1^2 + 2^2 + \cdots + k^2 + (k+1)^2$$ By the inductive hypothesis, replace the sum up to $k^2$: $$= \frac{k(k+1)(2k+1)}{6} + (k+1)^2$$ Factor out $(k+1)$: $$= \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6} = \frac{(k+1)\left[k(2k+1) + 6(k+1)\right]}{6}$$ Simplify inside the bracket: $$k(2k+1) + 6(k+1) = 2k^2 + k + 6k + 6 = 2k^2 + 7k + 6$$ Factor the quadratic: $$2k^2 + 7k + 6 = (2k+3)(k+2)$$ So the expression becomes: $$\frac{(k+1)(k+2)(2k+3)}{6}$$ Rewrite $2k+3$ as $2(k+1)+1$: $$= \frac{(k+1)(k+2)(2(k+1)+1)}{6}$$ This matches the right side of $P(k+1)$, completing the inductive step. **Final conclusion:** By mathematical induction, the formula $$1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$ holds for all positive integers $n$.