`((n+2)!)/((n-1)!)` is divisible by

Options

A. 6
B. 11
C. 24
D. 26

Answer

To determine the divisibility of \(\frac{(n+2)!}{(n-1)!}\), we can simplify the expression and analyze it step by step.

### Step 1: Simplify the Expression
We start with the expression:
\[ \frac{(n+2)!}{(n-1)!} \]
Using the factorial definition, we can expand \((n+2)!\):
\[ (n+2)! = (n+2)(n+1)n(n-1)! \]
Thus, we can rewrite the expression as:
\[ \frac{(n+2)(n+1)n(n-1)!}{(n-1)!} = (n+2)(n+1)n \]

### Step 2: Analyze the Product
Now, we need to analyze the product:
\[ (n+2)(n+1)n \]
This product consists of three consecutive integers: \(n\), \(n+1\), and \(n+2\).

### Step 3: Check Divisibility by 6
To determine the divisibility by 6, we note that:
- Among any three consecutive integers, at least one of them is divisible by 2 (even).
- Among any three consecutive integers, at least one of them is divisible by 3.

Thus, the product \((n+2)(n+1)n\) is divisible by both 2 and 3, which means it is divisible by:
\[ 2 \times 3 = 6 \]

### Step 4: Conclusion
Since we have shown that \((n+2)(n+1)n\) is divisible by 6 for any integer \(n\), we conclude that:
\[ \frac{(n+2)!}{(n-1)!} \text{ is divisible by } 6. \]

### Final Answer
The expression \(\frac{(n+2)!}{(n-1)!}\) is divisible by **6**.

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