We will prove the statement:

$1 + 3 + 5 + \dots + (2n - 1) = n^2$

for all $n \in \mathbb{N}$ (positive integers), by mathematical induction.

Step 1: Base Case

First, we verify the base case for $n = 1$.

When $n = 1$, the left-hand side of the equation is just the first odd number:

$1 = 1^2$

The base case holds since both sides are equal.

Step 2: Inductive Hypothesis

Now, assume that the formula is true for some arbitrary positive integer $k$. That is, assume:

$1 + 3 + 5 + \dots + (2k - 1) = k^2$

This is the inductive hypothesis.

Step 3: Inductive Step

We need to prove that the formula holds for $n = k + 1$. In other words, we need to show that:

$1 + 3 + 5 + \dots + (2k - 1) + [2(k + 1) - 1] = (k + 1)^2$

Using the inductive hypothesis, the left-hand side becomes:

$k^2 + [2(k + 1) - 1]$

Simplify the new term:

$2(k + 1) - 1 = 2k + 2 - 1 = 2k + 1$

So, the left-hand side becomes:

$k^2 + (2k + 1)$

Now, simplify the right-hand side:

$k^2 + 2k + 1 = (k + 1)^2$

Thus, the formula holds for $n = k + 1$ as well.

Conclusion:

Since the base case holds, and we have shown that if the formula is true for $n = k$, it is also true for $n = k + 1$, by the principle of mathematical induction, the formula is true for all $n \in \mathbb{N}$.

$1 + 3 + 5 + \dots + (2n - 1) = n^2 \quad \text{for all} \, n \in \mathbb{N}.$

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Would you like further details on any of these steps? Here are some related questions for deeper understanding:

1. What is the principle of mathematical induction?
2. Can we generalize this proof to other types of sequences?
3. How does mathematical induction differ from strong induction?
4. Can this formula be derived from the sum of an arithmetic series?
5. What are some practical applications of this identity in mathematics?

Tip: When performing induction, ensure your base case is clearly valid, as it serves as the foundation for the entire proof.