Question

If ${\Delta }_r=\left|\begin{array}{ccc}r& 2r-1& 3r-2\\ \frac{n}{2}& n-1& a\\ \frac{1}{2}n(n-1)& (n-1{)}^2& \frac{1}{2}(n-1)(3n-4)\end{array}\right|$, then the value of $\sum _{r=1}^{n-1}{\Delta }_r$ :

• A. Depends only on a
• B. Depends only on n
• C. Depends both on a and n
• D. Is independent of both a and n.

Answer 1

The correct option is D Is independent of both a and n.

${\sum }_{r=1}^{n-1}{\Delta }_r={\Delta }_1+{\Delta }_2+{\Delta }_{3r}+.......+{\Delta }_{n-1}$

$=\left|\begin{array}{ccc}1& 1& 1\\ \frac{n}{2}& n-1& a\\ \frac{n(n-1)}{2}& {(n-1)}^2& \frac{(n-1)(3n+4)}{2}\end{array}\right|+\left|\begin{array}{ccc}2& 3& 4\\ \frac{n}{2}& n-1& a\\ \frac{n(n-1)}{2}& {(n-1)}^2& \frac{(n-1)(3n+4)}{2}\end{array}\right|+.........+\left|\begin{array}{ccc}n-1& 2n-3& 3n-5\\ \frac{n}{2}& n-1& a\\ \frac{n(n-1)}{2}& {(n-1)}^2& \frac{(n-1)(3n-4)}{2}\end{array}\right|$

$=\left|\begin{array}{ccc}1+2+3+....+(n-1)& 1+3+5+....+(2n-3)& 1+4+7+....(3n-5)\\ \frac{n}{2}(n-1)& (n-1)(n-1)& a(n-1)\\ \frac{n(n-1)}{2}& {(n-1)}^2& \frac{(n-1)(3n-4)}{2}\end{array}\right|$
$=\left|\begin{array}{ccc}\frac{n(n-1)}{2}& {(n-1)}^2& \frac{(n-1)(3n-4)}{2}\\ \frac{n}{2}(n-1)& (n-1)(n-1)& a(n-1)\\ \frac{n(n-1)}{2}& {(n-1)}^2& \frac{(n-1)(3n-4)}{2}\end{array}\right|$

$=0$

So,${\sum }_{r=1}^{n-1}{\Delta }_r$ is independent of both a and n. .