If \( \frac{(n+2)!-(n+1)!}{n!}=100 \), what is the value of (n)?

Correct answer: 9

Using \((n+2)!=(n+2)(n+1)!\), the numerator becomes \((n+1)![(n+2)-1]=(n+1)!(n+1)\). On dividing by \(n!\), we get \((n+1)^2=100\). Since \(n\) must be a non-negative integer for factorial notation, \(n+1=10\), so \(n=9\). For option 10, the value would be \(121\), not 100. Exam tip: In factorial expressions, factor out the smaller factorial first for quick simplification.