To prove the statement $\forall n \in \mathbb{N}: 1 + 3 + 5 + \cdots + (2n-1) = n^2$ using mathematical induction, we will follow these steps:

1. Base Case: Verify the statement for $n = 1$.
2. Inductive Hypothesis: Assume the statement is true for some $n = k$, i.e., $1 + 3 + 5 + \cdots + (2k-1) = k^2$.
3. Inductive Step: Prove that the statement is true for $n = k + 1$, i.e., $1 + 3 + 5 + \cdots + (2k-1) + (2(k+1)-1) = (k+1)^2$.

Step-by-Step Proof:

1. Base Case

For $n = 1$: $1 = 1^2$ The statement is true for $n = 1$.

2. Inductive Hypothesis

Assume that for some $n = k$, the statement is true: $1 + 3 + 5 + \cdots + (2k-1) = k^2$

3. Inductive Step

We need to prove that the statement holds for $n = k + 1$: $1 + 3 + 5 + \cdots + (2k-1) + [2(k+1)-1] = (k+1)^2$

Start with the left-hand side of the equation for $n = k+1$: $1 + 3 + 5 + \cdots + (2k-1) + (2k+1)$

Using the inductive hypothesis $1 + 3 + 5 + \cdots + (2k-1) = k^2$, we can substitute: $k^2 + (2k + 1)$

Now, simplify the expression: $k^2 + 2k + 1 = (k + 1)^2$

Thus, we have shown that: $1 + 3 + 5 + \cdots + (2k-1) + (2k+1) = (k+1)^2$

This completes the inductive step.

Conclusion

By the principle of mathematical induction, we have shown that for all $n \in \mathbb{N}$: $1 + 3 + 5 + \cdots + (2n-1) = n^2$

This completes the proof.

Do you have any questions or need further details on any step?

Relative Questions:

1. What is the principle of mathematical induction?
2. How do you verify the base case in mathematical induction?
3. What is the significance of the inductive hypothesis in an induction proof?
4. How can you simplify expressions involving sums of arithmetic sequences?
5. What are some common pitfalls when performing mathematical induction?
6. Can you apply mathematical induction to prove inequalities?
7. How does mathematical induction differ from strong induction?
8. What are some real-world applications of mathematical induction?

Tip:

When working on proofs, especially using induction, carefully check each step and ensure your base case and inductive step are clearly and rigorously demonstrated.