Question

If $n\in N$ and
${\Delta }_n=\left|\begin{array}{ccc}n!& (n+1)!& (n+2)!\\ (n+1)!& (n+2)!& (n+3)!\\ (n+2)!& (n+3)!& (n+4)!\end{array}\right|$
then $\underset{n\to \infty }{lim}\frac{(3n^3-5){\Delta }_n}{{\Delta }_{n+1}}$

• A. $\frac{3}{2}$
• B. $\frac{5}{2}$
• C. $-\frac{5}{2}$
• D. $3$

Answer 1

Answer: (D)

$3$

taking $n!$ common from $R_1,(n+1)!from{R}_{2}and(n+3)!from{R}_3$ we obtain

${\Delta }_n=n!(n+1)!(n+2)!\Delta$

where

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

Using $R_3\to R_3-R_{2}and{R}_2\to R_2-R_1$ we get

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

$=\left|\begin{array}{cc}1& 2n+4\\ 1& 2n+6\end{array}\right|=2$

Thus, ${\Delta }_n=2n!(n+1)!(n+2)!$

$\Rightarrow {\Delta }_{n+1}=2(n+1)!(n+2)!(n+3)!$

Now,
$\frac{(3n^3-5){\Delta }_n}{{\Delta }_{n+1}}=\frac{(3n^3-5)}{(n+1)(n+2)(n+3)}$

$\Rightarrow \underset{n\to \infty }{lim}\frac{(3n^3-5){\Delta }_n}{{\Delta }_{n+1}}=\underset{n\to \infty }{lim}\frac{(3-\frac{5}{n^3})}{(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})}$

$=3$

Hence, option D is correct.