1. **Problem Statement:** Find the value of $n$ for the expression $$\frac{(n+2)!}{(n-1)!}.$$ 2. **Recall the factorial definition:** $$n! = n \times (n-1) \times (n-2) \times \cdots \times 1$$ and for any integers $a > b$, $$\frac{a!}{b!} = a \times (a-1) \times \cdots \times (b+1).$$ 3. **Rewrite the expression:** $$\frac{(n+2)!}{(n-1)!} = (n+2) \times (n+1) \times n.$$ This is because $(n+2)! = (n+2)(n+1)n(n-1)!$, so dividing by $(n-1)!$ leaves the product of the three terms. 4. **Final simplified form:** $$\frac{(n+2)!}{(n-1)!} = n(n+1)(n+2).$$ 5. **Interpretation:** The value of the expression depends on $n$, and it equals the product of three consecutive integers starting from $n$. Since the problem only asks to find the value of the expression in terms of $n$, the simplified form is the answer.