Question

The expression \( \frac{n(n + 1)^2(n+2)}{12} \) for all \( n \in \mathbb{N} \) is always:

• A. \( \frac{(n+2)^{4}}{12} \)
• B. \( \frac{n^{4}}{12} \)
• C. an integer
• D. \( \frac{(2n+1)n}{3} \)

Hint:

For divisibility questions involving consecutive integers, always split the problem into checking factors of primes (like 2, 3, 4, 6, 12). This makes proofs much faster in exams.

Answer 1

The Correct Option is C

Concept: We need to check whether \[ E_n = \frac{n(n+1)^2(n+2)}{12} \] is always an integer. This reduces to proving that the numerator is always divisible by \(12 = 3 \times 4\).

Step 1: Check divisibility by 3.
Among three consecutive integers \(n, n+1, n+2\), one is always divisible by 3. Since the numerator contains all three consecutive numbers, it is always divisible by 3.

Step 2: Check divisibility by 4.
We prove that the numerator always contains at least \(2^2\).

• If \(n+1\) is even, then \((n+1)^2\) contributes at least \(4\).

• If \(n+1\) is odd, then \(n\) and \(n+2\) are both even, contributing at least \(2 \times 2 = 4\).
Thus, in both cases, the expression is divisible by 4.

Step 3: Conclusion.
Since the numerator is divisible by both 3 and 4, it is divisible by \(12\). Hence, \[ \frac{n(n+1)^2(n+2)}{12} \in \mathbb{Z} \quad \forall n \in \mathbb{N}. \]