If \(\frac{(n+2)!-(n+1)!}{n!}=36\), what is the value of (n)?

Correct answer: 5

Write \((n+2)!=(n+2)(n+1)!\). The numerator becomes \((n+1)!\big[(n+2)-1\big]=(n+1)!(n+1)\). Since \((n+1)!=(n+1)n!\), the given expression simplifies to \((n+1)^2\). Thus, \((n+1)^2=36\), giving \(n+1=6\) and \(n=5\). The other algebraic value, \(n=-7\), is not valid because factorials of negative integers are undefined. Exam tip: Express factorial terms in terms of the smallest factorial before cancelling.