If \(\frac{(n+1)!}{(n-1)!}=72\), what is the value of (n)?

Correct answer: 8

Since \((n+1)!=(n+1)n(n-1)!\), the factor \((n-1)!\) cancels, so the given expression becomes \(n(n+1)\). Thus, \(n(n+1)=72\). The consecutive integers whose product is 72 are 8 and 9; hence, \(n=8\). The closest option, 7, gives \(7\times8=56\), not 72. Exam tip: expand the larger factorial up to the smaller factorial and then cancel common factors.