Question

If ${△}_r=\left|\begin{array}{ccc}2r-1& {}^{m}C_r& 1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m\right)& {\sin }^2\left(m\right)& {\sin }^2\left(m+1\right)\end{array}\right|$ then the value $\sum _{r=0}^m{△}_r=$

• A. $1$
• B. $0$
• C. $2$
• D. None of these

Answer 1

The correct option is B

$0$

Explanation for the correct option:

Find the value of $\sum _{r=0}^m{△}_r$:

Given,

${△}_r=\left|\begin{array}{ccc}2r-1& {}^{m}C_r& 1\\ m^2-1& 2^m& m+1\\ {\sin }^2\left(m\right)& {\sin }^2\left(m\right)& {\sin }^2\left(m+1\right)\end{array}\right|$

Now the value of

$\begin{array}{rcl}\sum _{r=0}^M{△}_r& =& \left|\begin{array}{ccc}\sum _{r=0}^{M}2r-1& \sum _{r=0}^M{}^{M}C_r& \sum _{r=0}^{M}1\\ M^2-1& 2^M& M+1\\ {\sin }^2\left(M\right)& {\sin }^2\left(M\right)& {\sin }^2\left(M+1\right)\end{array}\right|\\ & =& \left|\begin{array}{ccc}{M}^2-1& 2^M& M+1\\ M^2-1& 2^M& M+1\\ {\sin }^2\left(M\right)& {\sin }^2\left(M\right)& {\sin }^2\left(M+1\right)\end{array}\right|\end{array}$

$\begin{array}{rcl}\underset{\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{r}\mathbf{=}\mathbf{0}}{\overset{\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{M}}{\mathbf{\therefore }\mathbf{\sum }}}{\mathbf{△}}_{\mathbf{r}}\mathbf{}& \mathbf{=}& \mathbf{}\mathbf{0}\end{array}$ $\left[\mathbf{\because }{\mathit{R}}_{\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{R}}_{\mathbf{2}}\right]$

Hence, Option ‘B’ is Correct.