The formula for the sum of the first $n$ odd numbers is:

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

Proof by Mathematical Induction:

Step 1: Base Case

For $n = 1$, the sum is: $S_1 = 1$ The formula gives: $n^2 = 1^2 = 1$ Thus, the base case holds.

Step 2: Inductive Hypothesis

Assume the formula holds for some $n = k$: $S_k = 1 + 3 + 5 + \cdots + (2k - 1) = k^2$

Step 3: Inductive Step

We need to prove that the formula holds for $n = k + 1$, i.e., $S_{k+1} = 1 + 3 + 5 + \cdots + (2k - 1) + (2(k+1) - 1) = (k+1)^2$

From the inductive hypothesis: $S_k = k^2$ Add the next term, $2(k+1) - 1$, to both sides: $S_{k+1} = k^2 + (2(k+1) - 1)$ Simplify the added term: $S_{k+1} = k^2 + 2k + 2 - 1 = k^2 + 2k + 1$ Factorize the right-hand side: $S_{k+1} = (k + 1)^2$

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

Conclusion:

By the principle of mathematical induction, the formula $S_n = n^2$ is true for all $n \geq 1$.

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Would you like further details or have any questions?

Here are 5 related questions:

1. How can you derive the formula $S_n = n^2$ from scratch?
2. What is the formula for the sum of the first $n$ even numbers?
3. How does the proof for sums of arithmetic series differ from this?
4. Can you find the sum of the first $n$ odd numbers squared ($1^2 + 3^2 + 5^2 + \cdots$)?
5. How does this sum relate to the area of a square geometrically?

Tip: Always start with small examples to verify your formula before proving it rigorously!