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 $[\frac{D}{(n!{)}^3}-4]$ is___

• A. divisible by n.
• B. divisible by n+1
• C. divisible by n+2
• D. divisible by n+3

Answer 1

The correct option is A divisible by n.

Given, $D=\left|\begin{array}{ccc}n!& (n+1)!& (n+2)!\\ (n+1)!& (n+2)!& (n+3)!\\ (n+2)!& (n+3)!& (n+4)!\end{array}\right|$
Taking n!, (n+1)!and (n+2)! common from $R_1,R_{2}and{R}_3$, respectively.
$\therefore D=n!(n+1)!(n+2)!\left|\begin{array}{ccc}1& (n+1)& (n+1)(n+2)\\ 1& (n+2)& (n+2)(n+3)\\ 1& (n+3)& (n+3)(n+4)\end{array}\right|$
$Applying{R}_2\to R_2-R_{1}and{R}_3\to R_3-R_2$, we get
$D=n!(n+1)!(n+2)!\left|\begin{array}{ccc}1& (n+1)& (n+1)(n+2)\\ 0& 1& 2n+4\\ 0& 1& 2n+6\end{array}\right|$
Expanding along $C_1$, we get
D = (n!) (n+ 1)! (n + 2)! [(2n + 6) - (2n + 4)]
D = (n!) (n + 1)! (n + 2)! [2]
On dividing both side by $(n!{)}^3$
$\Rightarrow \frac{D}{(n!{)}^3}=\frac{(n!)(n!)(n+1)(n!)(n+1)(n+2)2}{(n!{)}^3}\Rightarrow \frac{D}{(n!{)}^3}=2(n+1)(n+1)(n+2)\Rightarrow \frac{D}{(n!{)}^3}=2(n^3+4n^2+5n+2)=2n(n^2+4n+5)+4\Rightarrow \frac{D}{(n!{)}^3}-4=2n(n^2+4n+5)$
which shows that $[\frac{D}{(n!{)}^3}-4]$ is divisible by n.