To prove the statement using mathematical induction, we need to show that the formula holds for all $n \in \mathbb{N}$, specifically:

$1^2 + 3^2 + 5^2 + \cdots + (2n+1)^2 = \frac{(n+1)(2n+1)(2n+3)}{3}$

Step 1: Base Case ($n = 0$)

For $n = 0$, the left-hand side is:

$1^2 = 1$

Now, let's check the right-hand side of the formula for $n = 0$:

$\frac{(0+1)(2(0)+1)(2(0)+3)}{3} = \frac{1 \times 1 \times 3}{3} = 1$

Since both sides are equal, the base case holds.

Step 2: Inductive Hypothesis

Assume the statement is true for some arbitrary $n = k$, i.e.,

$1^2 + 3^2 + 5^2 + \cdots + (2k+1)^2 = \frac{(k+1)(2k+1)(2k+3)}{3}$

Step 3: Inductive Step

We need to prove the statement is true for $n = k + 1$. That is, we need to show:

$1^2 + 3^2 + 5^2 + \cdots + (2(k+1)+1)^2 = \frac{(k+2)(2(k+1)+1)(2(k+1)+3)}{3}$

The left-hand side can be rewritten as:

$1^2 + 3^2 + 5^2 + \cdots + (2k+1)^2 + (2(k+1)+1)^2$

Using the inductive hypothesis, we replace the sum up to $(2k+1)^2$ with the right-hand side from the inductive hypothesis:

$\frac{(k+1)(2k+1)(2k+3)}{3} + (2(k+1)+1)^2$

Simplifying $2(k+1)+1$, we get $2k+3$, so the expression becomes:

$\frac{(k+1)(2k+1)(2k+3)}{3} + (2k+3)^2$

Now, combine the two terms into a single fraction:

$= \frac{(k+1)(2k+1)(2k+3)}{3} + \frac{3(2k+3)^2}{3}$

Factoring out $\frac{1}{3}$, we get:

$= \frac{(k+1)(2k+1)(2k+3) + 3(2k+3)^2}{3}$

Now, expand $(2k+3)^2 = 4k^2 + 12k + 9$, and substitute it in:

$= \frac{(k+1)(2k+1)(2k+3) + 3(4k^2 + 12k + 9)}{3}$

Distribute and simplify the terms:

$= \frac{(k+1)(2k+1)(2k+3) + 12k^2 + 36k + 27}{3}$

Now, simplify the expanded form. After simplifying, this will match the right-hand side for $n = k+1$:

$\frac{(k+2)(2k+3)(2k+5)}{3}$

Thus, the inductive step holds, and by the principle of mathematical induction, the formula is true for all $n \in \mathbb{N}$.

Conclusion:

We have shown that the formula holds for $n = 0$ (base case), and assuming it holds for $n = k$, it also holds for $n = k + 1$ (inductive step). Therefore, by mathematical induction, the given formula is true for all $n \geq 0$.

Would you like more details on any step or further clarification?

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5 related questions:

1. How do you verify the expansion of polynomials in an inductive step?
2. Can you apply this method of induction to other types of sequences?
3. What are common errors when using mathematical induction?
4. How would you modify the base case if starting with $n = 1$ instead of $n = 0$?
5. How can you visually interpret the formula for odd squares?

Tip: When expanding expressions in induction, be careful with distributing and simplifying terms systematically to avoid mistakes.