Question

For a fixed positive integer $n$, if$△=\left|\begin{array}{ccc}n!& (n+1)!& (n+2)!\\ (n+1)!& (n+2)!& (n+3)!\\ (n+2)!& (n+3)!& (n+4)!\end{array}\right|$Then show that $[\frac{△}{{(n!)}^3}-4]$ is divisible by $n$.

Answer 1

$\Delta =\left|\begin{array}{ccc}n!& \left(n+1\right)!& \left(n+2\right)!\\ \left(n+1\right)!& \left(n+2\right)!& \left(n+3\right)!\\ \left(n+2\right)!& \left(n+3\right)!& \left(n+4\right)!\end{array}\right|$

$\Delta =\left|\begin{array}{ccc}n!& \left(n+1\right)n!& \left(n+2\right)\left(n+1\right)n!\\ \left(n+1\right)!& \left(n+2\right)\left(n+1\right)!& \left(n+3\right)\left(n+2\right)\left(n+1\right)!\\ \left(n+2\right)!& \left(n+3\right)\left(n+2\right)!& \left(n+4\right)\left(n+3\right)\left(n+2\right)!\end{array}\right|$

Taking out common $n!,(n+1)!,(n+2)!$ from $R_1,R_2,R_3$ respectively.

$=n!\left(n+1\right)!\left(n+2\right)!\left|\begin{array}{ccc}1& \left(n+1\right)& \left(n+2\right)\left(n+1\right)\\ 1& \left(n+2\right)& \left(n+3\right)\left(n+2\right)\\ 1& \left(n+3\right)& \left(n+4\right)\left(n+3\right)\end{array}\right|$

$R_2\to R_2-R_1$

$R_3\to R_3-R_1$

$=n!\left(n+1\right)!\left(n+2\right)!\left|\begin{array}{ccc}1& \left(n+1\right)& \left(n+2\right)\left(n+1\right)\\ 0& 1& 2\left(n+2\right)\\ 0& 2& \left(n+3\right)\left(n+4\right)-\left(n+1\right)\left(n+2\right)\end{array}\right|$

Expanding along $C_1$

$=n!\left(n+1\right)!\left(n+2\right)!\left[\left(n+3\right)\left(n+4\right)-\left(n+1\right)\left(n+2\right)-4\left(n+2\right)\right]$

$=n!\left(n+1\right)!\left(n+2\right)!\left[2\right]$

$=2n!\left(n+1\right)n!\left(n+2\right)\left(n+1\right)n!$

$\frac{\Delta }{{(n!)}^3}=2\left(n+1\right)\left(n+2\right)\left(n+1\right)$

$\frac{\Delta }{{(n!)}^3}=2\left[n^3+4n^2+5n+2\right]$

$\frac{\Delta }{{(n!)}^3}=2n\left[n^2+4n+5\right]+4$

$\frac{\Delta }{{(n!)}^3}-4=2n\left[n^2+4n+5\right]$

$\frac{\Delta }{{(n!)}^3}-4$ is divisible by $n$

proved.