To prove that the expression

$\frac{n(n+1)(n+2)}{6}$

is an integer for $n \geq 1$ using the division algorithm, we can approach this by examining the product $n(n+1)(n+2)$ and showing that it is divisible by 6.

Step-by-Step Proof:

1. Analyze the Product $n(n+1)(n+2)$:

   • The three consecutive integers $n$, $n+1$, and $n+2$ contain at least one multiple of 2 and one multiple of 3 (by properties of consecutive integers).
   • This ensures that $n(n+1)(n+2)$ is divisible by both 2 and 3, making it divisible by 6.

2. Divisibility by 2:

   • In any set of three consecutive numbers, at least one number is even, guaranteeing that the product $n(n+1)(n+2)$ is divisible by 2.

3. Divisibility by 3:

   • Similarly, in any set of three consecutive numbers, one of the numbers is divisible by 3. Therefore, the product $n(n+1)(n+2)$ is also divisible by 3.

4. Conclusion on Divisibility by 6:

   • Since $n(n+1)(n+2)$ is divisible by both 2 and 3, it must be divisible by 6.

5. Resulting Integer:

   • Dividing $n(n+1)(n+2)$ by 6 gives an integer, as required.

Thus, we have shown that $\frac{n(n+1)(n+2)}{6}$ is indeed an integer for all $n \geq 1$.

Would you like more details, or do you have any questions about the proof?

Here are five related questions for further exploration:

1. What are the general properties of consecutive integers regarding divisibility?
2. How does this proof change if we use a different divisor, say 4 or 8?
3. What other methods, besides the division algorithm, can we use to show that an expression is an integer?
4. Can this method be extended to prove divisibility for expressions involving more consecutive numbers?
5. What is the significance of the product of three consecutive integers in combinatorial problems?

Tip: When proving divisibility properties, breaking down products involving consecutive numbers is a helpful strategy due to their predictable divisibility patterns.