Question

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

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

Answer 1

Answer: (D)

$-64$

Take $(n-1)!$, $(n+1)!$, $(n+3)!$ common from $R_1$, $R_2$, $R_3$ respectively

$\therefore \frac{D}{(n-1)!(n+1)!(n+3)!}$

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

Making two zeros in column $I$ by applying $R_3-R_2$ ad $R_2-R_1$, we have

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

or $E=\left|\begin{array}{cc}4n+6& 8\\ 4n+14& 8\end{array}\right|$ by $C_2-C_1$

$=8[4n+6-4n-14]=8[-8]=-64\Rightarrow \left(d\right)$