Question

Let $m$ be a positive integer and ${\Delta }_r=\left|\begin{array}{ccc}2r-1& {}^m{C}_r& 1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m^2\right)& {\sin }^2\left(m\right)& {\sin }^2(m+1)\end{array}\right|(0\le r\le m).$ Then the value of ${\sum }_{r=0}^m{\Delta }_r$ is given by

Answer 1

$\begin{array}{c}{\Delta }_r=\left(\begin{array}{ccc}2r-1& {}^m{C}_r& 1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m^2\right)& {\sin }^{2}m& {\sin }^2(m+1)\end{array}\right)\\ sincemisconstant,\\ {\sum }_{r=0}^m{\Delta }_r=\left(\begin{array}{ccc}2{\sum }_{r=0}^{m}r-{\sum }_{r=0}^{m}1& {\sum }_{r=0}^m{}^m{C}_r& {\sum }_{r=0}^{m}1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m^2\right)& {\sin }^2\left(m\right)& {\sin }^2(m+1)\end{array}\right)\\ =\left(\begin{array}{ccc}2\left\{\frac{m(m+1)}{2}\right\}-(m+1)& 2^m& m+1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m^2\right)& {\sin }^{2}m& {\sin }^2(m+1)\end{array}\right)\\ =\left(\begin{array}{ccc}{m}^2-1& 2^m& m+1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m^2\right)& {\sin }^{2}m& {\sin }^2(m+1)\end{array}\right)\\ =0\\ As{R}_{1=}{R}_2\\ Hence\Delta =0\end{array}$