If \( \frac{(n+2)!}{n!}+\frac{n!}{(n-2)!}=114 \), what is the value of (n)?

Correct answer: 7

\(\frac{(n+2)!}{n!}=(n+2)(n+1)\) and \(\frac{n!}{(n-2)!}=n(n-1)\). Thus, \((n+2)(n+1)+n(n-1)=114\). Simplifying gives \(2n^2+2n+2=114\), or \(n^2+n-56=0\). Hence, \((n+8)(n-7)=0\), so \(n=7\) or \(-8\). Since the factorial expression requires \(n\ge2\), the valid value is \(n=7\). Substituting \(n=8\) gives 146 on the left-hand side, so it is a close but incorrect option. Exam tip: cancel factorials first by expanding only the required consecutive factors.