Let's solve Problem 1 step by step using mathematical induction.

Problem Statement:

We need to prove that:

$1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$

for all positive integers $n$.

---

Part (a): What is the statement $P(1)$?

The statement $P(1)$ means substituting $n = 1$ into the formula: $1^2 = \frac{1(1+1)(2 \cdot 1 + 1)}{6}$

Simplify the right-hand side: $\frac{1(2)(3)}{6} = \frac{6}{6} = 1$

So, $P(1)$ is true.

---

Part (b): Show that $P(1)$ is true, completing the basis step.

As shown above, $P(1)$ holds because both the left-hand side and right-hand side equal 1. This completes the basis step.

---

Part (c): What is the inductive hypothesis?

The inductive hypothesis is to assume that $P(k)$ is true for some arbitrary positive integer $k$. That is: $1^2 + 2^2 + \dots + k^2 = \frac{k(k+1)(2k+1)}{6}$

---

Part (d): What do you need to prove in the inductive step?

We need to prove that $P(k+1)$ is true, i.e., that: $1^2 + 2^2 + \dots + k^2 + (k+1)^2 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$

---

Part (e): Complete the inductive step.

Start with the left-hand side of $P(k+1)$: $1^2 + 2^2 + \dots + k^2 + (k+1)^2$

Using the inductive hypothesis, substitute $1^2 + 2^2 + \dots + k^2 = \frac{k(k+1)(2k+1)}{6}$: $\frac{k(k+1)(2k+1)}{6} + (k+1)^2$

Factor $(k+1)$ out of the expression: $\frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6}$ $= \frac{(k+1)}{6} \left[ k(2k+1) + 6(k+1) \right]$

Simplify the terms inside the brackets: $k(2k+1) + 6(k+1) = 2k^2 + k + 6k + 6 = 2k^2 + 7k + 6$

Factor $2k^2 + 7k + 6$: $2k^2 + 7k + 6 = (k+2)(2k+3)$

Substitute back: $\frac{(k+1)}{6} \left[ (k+2)(2k+3) \right]$

Simplify: $\frac{(k+1)(k+2)(2k+3)}{6}$

This matches the right-hand side of $P(k+1)$. Thus, $P(k+1)$ is true.

---

Conclusion:

By the principle of mathematical induction, the statement: $1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$ is true for all positive integers $n$.

---

Would you like a detailed breakdown of any specific step or clarification on the inductive hypothesis?

---

Related questions:

1. How does the proof change if the formula changes to the sum of cubes?
2. Can we use induction for non-integer sequences?
3. How can this formula be derived without induction?
4. What happens if $n = 0$ is included in the sequence?
5. How is this formula applied in real-world problems?

Tip: Mathematical induction is a systematic process—focus on proving the base case and the inductive step rigorously.