Question

Find $n$, if $\left(n+1\right)!=12\times \left(n-1\right)!$

Answer 1

We have,

$(n+1)!=12\times (n-1)!$

$(n+1)\times (n+1-1)\times (n+1-2)!=12\times (n-1)!$

$(n+1)\times n\times (n-1)!=12\times (n-1)!$

$(n+1)\times n=12$

$n^2+n-12=0$

$n^2+4n-3n-12=0$

$n(n+4)-3(n+4)=0$

$(n+4)(n-3)=0$

$n=3,-4$

Hence, this is the answer.