Question

If $(n+1)!=30\cdot (n-1)!$, then the value of $n$ is

Answer 1

Given : $(n+1)!=30\cdot (n-1)!$
$\Rightarrow (n+1)n(n-1)!=30\cdot (n-1)!$
$\Rightarrow n(n+1)=30$
$\Rightarrow n^2+n-30=0$
$\Rightarrow (n+6)(n-5)=0$
$\Rightarrow n=-6,5$
$(\because n$ can't be negative$)$
$\Rightarrow n=5$