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Please prove the following statement with full induction, and explain every step


The task provided in the image asks for a proof of the following sum of squares formula using **mathematical induction**:

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

We need to prove this holds for all natural numbers \( n \).

### Steps for the Induction Proof:

#### 1. **Base Case (n = 1)**:
We first check if the formula is valid for \( n = 1 \).

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

This is correct, so the base case holds.

#### 2. **Induction Hypothesis**:
Assume that the formula holds for some arbitrary natural number \( n = k \). That is, we assume:

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

This is our **inductive hypothesis**.

#### 3. **Inductive Step**:
We now need to prove that the formula also holds for \( n = k+1 \). That is, we must show:

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

Using the inductive hypothesis, we can rewrite the left-hand side as:

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

Now, let's simplify the right-hand side of the equation we aim to prove. For \( n = k+1 \), the formula becomes:

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

Thus, we need to show that:

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

#### 4. **Simplification**:
Factor out \( (k+1) \) from both sides:

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

Now, simplify the term inside the parentheses:

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

So we now have:

\[
\frac{(k+1)}{6} \left( 2k^2 + 7k + 6 \right)
\]

This can be factored as:

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

Thus, both sides are equal, completing the inductive step.

#### 5. **Conclusion**:
By the principle of mathematical induction, the formula:

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

is true for all natural numbers \( n \).

---

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

Here are five follow-up questions to explore this topic further:

1. What is the next number in the sequence \( 1^2 + 2^2 + 3^2 + \dots + n^2 \) after \( n = 4 \)?
2. How can mathematical induction be applied to other summation formulas, such as the sum of cubes?
3. Can you use this method to find the sum of odd squares only, like \( 1^2 + 3^2 + 5^2 \)?
4. Why does factoring play such a crucial role in simplifying the inductive step?
5. What are other common series that can be proven using induction?

**Tip**: When dealing with induction, always ensure that the inductive hypothesis is clearly stated. It forms the bridge between the base case and the proof for \( n+1 \).

Prove by complete induction that 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6 for all natural numbers n.

Explore the proof of the sum of squares formula 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6 using mathematical induction. This detailed explanation covers each step of the inductive process, suitable for high school students in Grades 11-12.

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