Question

If ${\Delta }_r=\left|\begin{array}{ccc}2r-1& {}^m{C}_r& 1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m^2\right)& {\sin }^2\left(m\right)& {\sin }^2(m+1)\end{array}\right|,(0\le r\le m),$
then value of $\sum _{r=0}^m{\Delta }_r$ is

• A. $0$
• B. $m^2-1$
• C. $2^m$
• D. $2^m{\sin }^2\left(2^m\right)$

Answer 1

The correct option is A $0$

Using the sum property, we get $\sum _{r=0}^m{\Delta }_r=\left|\begin{array}{ccc}\sum _{r=0}^m(2r-1)& \sum _{r=0}^m{}^m{C}_r& \sum _{r=0}^{m}1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m^2\right)& {\sin }^2\left(m\right)& {\sin }^2(m+1)\end{array}\right|$
But,
$\sum _{r=0}^m(2r-1)=\sum _{r=1}^m(2r-1)-1=m^2-1\sum _{r=0}^m{}^m{C}_r=2^m\text{and}\sum _{r=0}^{m}1=m+1\therefore \sum _{r=0}^m={\Delta }_r=\left|\begin{array}{ccc}{m}^2-1& 2^m& m+1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m^2\right)& {\sin }^2\left(m\right)& {\sin }^2(m+1)\end{array}\right|=0$
$\{\because R_1\text{and}{R}_2\text{are similar}\}$