ĐK: n+1$\geq 3\rightarrow n\geq 2$
$\leftrightarrow \frac{(n+1)!}{(n-2)!}+\frac{(n+1)!}{2!(n-1)!}<14n+14$$\leftrightarrow 2(n+1)n(n-1)+(n+1)n-28n-28<0$
$\leftrightarrow 2n^{3}+n^{2}-29n-28<0$
$\leftrightarrow (n-4)(n+1)(2n+7)<0$. ta có n$\geq $2 nên $(n+1) và (2n+7) >0$
$\rightarrow (n+1)(2n+7)(n-4)<0 \leftrightarrow n-4<0 \leftrightarrow n<4$
KL: $2\leq n< 4$