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 ${\sum }_{r=0}^m{\Delta }_r$

• 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 $0$

${\Delta }_r=\left|\begin{array}{ccc}2r-1& {{}^{m}C}_r& 1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m^2\right)& {\sin }^2\left(m\right)& {\sin }^2(m+1)\end{array}\right|{\sum }_{r=0}^m{\Delta }_r=\left|\begin{array}{ccc}{\sum }_{r=0}^m(2r-1)& {\sum }_{r=0}^m{{}^{m}C}_r& {\sum }_{r=0}^{m}1\\ (m^2-1)& 2^m& m+1\\ {\sin }^2\left(m^2\right)& {\sin }^2\left(m\right)& {\sin }^2(m+1)\end{array}\right|=\left|\begin{array}{ccc}2\frac{m(m+1)}{2}-(m+1)& 2^m& m+1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m^2\right)& {\sin }^2\left(m\right)& {\sin }^2(m+1)\end{array}\right|=0$
Hence, option 'A' is correct.