Question

For a fixed positive integer $n$,let
$D=\left|\begin{array}{ccc}(n-1)!& (n+1)!& (n+3)!/n(n+1)\\ (n+1)!& (n+3)!& (n+5)!/(n+2)(n+3)\\ (n+3)!& (n+5)!& (n+7)!/(n+4)(n+5)!\end{array}\right|$ then $\frac{D}{(n-1)!(n+1)!(n+3)!}$ is equal to

• A. - 8
• B. - 16
• C. - 32
• D. - 64

Answer 1

Answer: (D)

- 64

From given taking $(n-1)!$ common from $R_1$, $(n+1)!$ from $R_2$ and $(n+3)!$ from $R_3$, we get

$D=(n-1)!(n+1)!(n+3)!D_1$

Where

$D_1=\left|\begin{array}{ccc}1& (n+1)n& (n+3)(n+2)\\ 1& (n+3)(n+2)& (n+5)(n+4)\\ 1& (n+5)(n+4)& (n+7)(n+6)\end{array}\right|$

Applying $R_3\to R_3-R_2$ and $R_2\to R_2-R_1$

$D_1=\left|\begin{array}{ccc}1& (n+1)n& (n+3)(n+2)\\ 0& 4n+6& 4n+14\\ 0& 4n+14& 4n+22\end{array}\right|=\left|\begin{array}{cc}4n+6& 4n+14\\ 4n+14& 4n+22\end{array}\right|=\left|\begin{array}{cc}4n+6& 8\\ 4n+14& 8\end{array}\right|$

Using $C_2\to C_2-C_1$

$D_1=\left|\begin{array}{cc}4n+6& 8\\ 8& 0\end{array}\right|=-64$

Thus $\frac{D}{(n-1)!(n+1)!(n+3)!}=-64$