If \( \frac{(n+1)!+n!}{(n-1)!}=63 \), what will be the value of (n)?

Correct answer: 7

Using \((n+1)!=(n+1)n!\) and \(n!=n(n-1)!\),
\[
\frac{(n+1)!+n!}{(n-1)!}=\frac{n![(n+1)+1]}{(n-1)!}=n(n+2).
\]
Thus, \(n(n+2)=63\), or \(n^2+2n-63=0\). This gives \((n-7)(n+9)=0\). Since a factorial is not defined for a negative integer, \(n=-9\) is invalid; hence \(n=7\). Exam tip: in factorial fractions, first express terms using the smallest factorial present, here \((n-1)!\), and then cancel.