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

Step 1: Understand the goal.
We must check what $E_n=\dfrac{n(n+1)^2(n+2)}{12}$ always is. Since $12 = 3 \times 4$, the question is really whether the top is always divisible by both 3 and 4.
Step 2: Look at the numbers on top.
The top contains $n$, $n+1$ and $n+2$, which are three numbers in a row, plus an extra $n+1$.
Step 3: Check divisibility by 3.
Among any three numbers in a row, one of them is always a multiple of 3. So the top is always divisible by 3.
Step 4: Check divisibility by 4, first case.
If $n+1$ is even, then $(n+1)^2$ holds two factors of 2, giving a 4. So the top is divisible by 4.
Step 5: Check divisibility by 4, second case.
If $n+1$ is odd, then $n$ and $n+2$ are both even. Two even numbers together give $2 \times 2 = 4$. So again the top is divisible by 4.
Step 6: Combine and conclude.
The top is divisible by 3 and by 4, so it is divisible by 12. The fraction is therefore a whole number for every natural $n$. \[ \boxed{\text{an integer}} \]