1. **Problem Statement:** Evaluate $$\frac{(n+2)!}{(n-1)!}$$ for a general $n$. 2. **Recall the factorial definition:** $$n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$$ 3. **Rewrite the expression:** $$\frac{(n+2)!}{(n-1)!} = \frac{(n+2)(n+1)n(n-1)!}{(n-1)!}$$ 4. **Cancel the common factorial term:** $$= (n+2)(n+1)n$$ 5. **Explanation:** Since $(n+2)! = (n+2)(n+1)n(n-1)!$, dividing by $(n-1)!$ cancels that term, leaving the product of the three consecutive integers. 6. **Final answer:** $$\boxed{(n+2)(n+1)n}$$ This expression gives the value of $$\frac{(n+2)!}{(n-1)!}$$ for any integer $n \geq 1$.