To prove that $n^2 - n$ is even using the sum $1 + 2 + 3 + \dots + (n-1)$, we can interpret $n^2 - n$ as the result of Proposition 4.2, which relates to sums of integers.

Proof:

1. Expression of $n^2 - n$: The expression $n^2 - n$ can be factored as: $n^2 - n = n(n - 1)$ This product is the multiplication of two consecutive integers $n$ and $n-1$.

2. Sum of first $n-1$ integers: The sum of the first $n-1$ positive integers is: $S = 1 + 2 + 3 + \dots + (n - 1)$ By the well-known formula for the sum of the first $m$ integers, we have: $S = \frac{(n - 1)n}{2}$

3. Relating $n^2 - n$ to the sum: We notice that the product $n(n - 1)$ appears in the numerator of the sum formula. Therefore: $n^2 - n = 2S$ This shows that $n^2 - n$ is exactly twice the sum of the first $n-1$ integers.

4. Conclusion: Since $n^2 - n = 2S$, it is clear that $n^2 - n$ is an even number, because it is twice the integer $S$. Hence, $n^2 - n$ is always even for any $n \in \mathbb{N}$.

Final Result:

$n^2 - n \text{ is even for any } n \in \mathbb{N}.$

Would you like any further details on this proof?

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Here are 5 related questions that build on this concept:

1. How do you prove that the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$?
2. Can you generalize the expression $n^2 - n$ for non-integer values of $n$?
3. What is the significance of the sum of consecutive integers in other areas of mathematics?
4. How does the parity (even or odd nature) of an integer affect the expression $n^2 - n$?
5. Can you explore similar results for products of three consecutive integers?

Tip: When dealing with sums or products of integers, always check for common factorizations that can simplify the proof, such as factoring out constants like 2 to check for evenness or oddness.