Question

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

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

Answer 1

The correct option is A $0$

By varying $r$,only $1st$ row determinant will vary and other $2$ rows will remain constant ,and as we can add determinants with two elements same (rows or columns ) and only one element varying (1st row here ) we can take the summation inside the $1st$ row

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

Now ,$\sum (2r-1)$ ${\sum }_r^C$ $\sum 1$

$=2(\sum r)\left(1\right)$ ${}^m{C}_0{+}^m{C}_1+....{+}^m{C}_m$ $=1+1+......+1$

$=2(0+1+.....+m)-(\sum 1)$ $2^m$ $m+1$

$=2\left(\frac{m(m+1)}{2}\right)-(\sum 1)$

$=m(m+1)-(\sum 1)$

$=m^2+m-(1++1+......+1)$

$=m^2+m-(m+1)$

$=m^2-1$

Therefore $1st$ and $2nd$ row of determinant are same and by property if determinate if two rows of it are same its value would be zero.