Question

For a fixed positive integer n, if $D=\left|\begin{array}{ccc}n!& (n+1)!& (n+2)!\\ (n+1)!& (n+2)!& (n+3)!\\ (n+2)!& (n+3)!& (n+4)!\end{array}\right|$ then show

$[\frac{D}{(n!{)}^3}-4]$ is divisible by n.

Answer 1

$D=\left|\begin{array}{ccc}n!& (n+1)!& (n+2)!\\ (n+1)!& (n+2)!& (n+3)!\\ (n+2)!& (n+3)!& (n+4)!\end{array}\right|$
Taking $(n!),(n+1)!$ and $(n+2)!$ common from $C_1,C_2$ and $C_3$, we get
$D=(n!)(n+1)!(n+2)!\left|\begin{array}{ccc}1& 1& 1\\ n+1& (n+2)& (n+3)\\ (n+2)(n+1)& (n+3)(n+2)& (n+4)(n+3)\end{array}\right|$
Applying $C_2\to C_2-C_1,C_3\to C_3-C_2$
$D=(n!)(n+1)!(n+2)!\left|\begin{array}{ccc}1& 0& 0\\ n+1& 1& 2\\ (n+2)(n+1)& 2(n+2)& 2(n+5)\end{array}\right|=(n!)(n+1)!(n+2)!(2n+2)$
Hence
$[\frac{D}{(n!{)}^3}-4]=[\frac{(n!)(n+1)!(n+2)!(2n+2)}{{(n!)}^3}-4]=[\frac{(n+1)(n+1)(n+2)(2n+2)}{1}-4]=(2(n^4+5n^3+9n^2+7n+2)-4)=2(n^4+5n^3+9n^2+7n)$