The theorem states:

For any $n \in \mathbb{Z}_{\geq 1}$,
$\sum_{i=1}^n (2i - 1) = n^2$

We will prove this by mathematical induction.

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

For $n = 1$, the left-hand side of the equation is: $\sum_{i=1}^1 (2i - 1) = 2(1) - 1 = 1.$

The right-hand side of the equation is: $n^2 = 1^2 = 1.$

Thus, the theorem holds for $n = 1$.
Base case verified.

Step 2: Inductive Hypothesis

Assume the theorem holds for some $n = k$, i.e., assume: $\sum_{i=1}^k (2i - 1) = k^2.$

Step 3: Inductive Step

We need to prove the theorem holds for $n = k + 1$, i.e., show: $\sum_{i=1}^{k+1} (2i - 1) = (k + 1)^2.$

Start with the left-hand side for $n = k + 1$: $\sum_{i=1}^{k+1} (2i - 1) = \left( \sum_{i=1}^k (2i - 1) \right) + (2(k+1) - 1).$

By the inductive hypothesis, we know: $\sum_{i=1}^k (2i - 1) = k^2.$

Substitute this into the equation: $\sum_{i=1}^{k+1} (2i - 1) = k^2 + (2(k+1) - 1).$

Simplify $2(k+1) - 1$: $\sum_{i=1}^{k+1} (2i - 1) = k^2 + 2k + 2 - 1 = k^2 + 2k + 1.$

Simplify further: $\sum_{i=1}^{k+1} (2i - 1) = (k + 1)^2.$

Thus, the theorem holds for $n = k + 1$.

Step 4: Conclusion

By the principle of mathematical induction, the theorem: $\sum_{i=1}^n (2i - 1) = n^2$ is true for all $n \in \mathbb{Z}_{\geq 1}$.

Would you like a detailed explanation of any specific step?

Relative Questions:

1. Why is the base case essential in mathematical induction?
2. How does the inductive hypothesis relate to the proof of $n = k + 1$?
3. Can you derive the general formula for summing odd numbers up to a given $n$?
4. What other types of problems can mathematical induction be used to solve?
5. How does this result connect to the concept of square numbers?

Tip:

In induction problems, always verify the hypothesis step carefully and connect it back to the desired equation for $n = k + 1$.