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To prove the given formula using mathematical induction, let's follow the standard process. The statement to prove is:

\[
S(n) = \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
\]

### Step 1: Base Case (for \( n = 1 \))

For \( n = 1 \), the left-hand side of the equation is:

\[
S(1) = \sum_{k=1}^{1} k^2 = 1^2 = 1
\]

Now, check the right-hand side for \( n = 1 \):

\[
\frac{1(1+1)(2(1)+1)}{6} = \frac{1(2)(3)}{6} = \frac{6}{6} = 1
\]

Thus, the base case holds true for \( n = 1 \).

### Step 2: Inductive Hypothesis

Assume the formula holds for some \( n = k \), i.e.,

\[
\sum_{i=1}^{k} i^2 = \frac{k(k+1)(2k+1)}{6}
\]

This is called the **inductive hypothesis**.

### Step 3: Inductive Step

We now prove that the formula holds for \( n = k+1 \). We need to show:

\[
\sum_{i=1}^{k+1} i^2 = \frac{(k+1)(k+2)(2(k+1)+1)}{6}
\]

Starting with the left-hand side:

\[
\sum_{i=1}^{k+1} i^2 = \left( \sum_{i=1}^{k} i^2 \right) + (k+1)^2
\]

Using the inductive hypothesis:

\[
\sum_{i=1}^{k+1} i^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2
\]

Factor out \( (k+1) \) from the right-hand side:

\[
\sum_{i=1}^{k+1} i^2 = \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6}
\]

Combine the terms over a common denominator:

\[
\sum_{i=1}^{k+1} i^2 = \frac{(k+1)\left(k(2k+1) + 6(k+1)\right)}{6}
\]

Simplify the expression inside the parentheses:

\[
k(2k+1) + 6(k+1) = 2k^2 + k + 6k + 6 = 2k^2 + 7k + 6
\]

Thus, we have:

\[
\sum_{i=1}^{k+1} i^2 = \frac{(k+1)(2k^2 + 7k + 6)}{6}
\]

Factor the quadratic expression:

\[
2k^2 + 7k + 6 = (k+2)(2k+3)
\]

So the equation becomes:

\[
\sum_{i=1}^{k+1} i^2 = \frac{(k+1)(k+2)(2k+3)}{6}
\]

This matches the right-hand side of the formula for \( n = k+1 \):

\[
\frac{(k+1)(k+2)(2(k+1)+1)}{6}
\]

### Conclusion

By the principle of mathematical induction, the formula

\[
\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
\]

is true for all \( n \geq 1 \).

---

Do you have any questions or would you like more details on any step?

Here are five related questions for further exploration:
1. How does mathematical induction differ from other proof techniques?
2. Can you use a similar method to prove the sum of cubes, i.e., \( \sum_{k=1}^{n} k^3 \)?
3. What is the relationship between the sum of squares formula and the formula for arithmetic sequences?
4. How would the proof change if the summation had different limits (e.g., starting from 0 instead of 1)?
5. How is the sum of squares related to the moments in statistics?

**Tip:** When performing induction proofs, it's essential to clearly distinguish between the base case, the inductive hypothesis, and the inductive step. Each step must logically follow the previous one.

Use induction to prove that the sum of the squares of the first n integers is equal to n(n+1)(2n+1)/6.

Learn how to use mathematical induction to prove the formula for the sum of the squares of the first n integers, S(n) = n(n+1)(2n+1)/6. This detailed step-by-step solution covers the base case, inductive hypothesis, and inductive step, providing a solid understanding of how to approach similar problems.

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