Question

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

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

Answer 1

Answer: (A)

Using the sum property we get $\sum _{r=0}^m{\Delta }_r$
$=\left|\begin{array}{ccc}\sum _{r=0}^m(2r-1)& \sum _{r=0}^m{C}_r& \sum _{r=0}^{m}1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m^2\right)& {\sin }^2(m)& {\sin }^2(m+1)\end{array}\right|$
But $\sum _{r=0}^m(2r-1)=\frac{1}{2}(m+1)(2m-1-1)=m^2-1$,
$\sum _{r=0}^m{}^m{C}_r=2^m$ and $\sum _{r=0}^{m}1=m+1$
Therefore, $\sum _{r=0}^m{\Delta }_r=\left|\begin{array}{ccc}{m}^2-1& 2^m& m+1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m^2\right)& {\sin }^2(m)& {\sin }^2(m+1)\end{array}\right|=0$