Question

If ${\Delta }_r=\left|\begin{array}{ccc}2r-1& {}^m{C}_r& 1\\ m^2-1& 2^m& m+1\\ si{n}^2\left(m^2\right)& si{n}^2\left(m\right)& si{n}^2(m+1)\end{array}\right|$, then the value of $\sum _{r=0}^m{\Delta }_r$, is

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

Answer 1

Answer: (B)

0

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

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

$=\left|\begin{array}{ccc}{m}^2-1& 2^m& m+1\\ m^2-1& 2^m& m+1\\ si{n}^2\left(m^2\right)& si{n}^2\left(m\right)& si{n}^2(m+1)\end{array}\right|$

$=0$ ($\because$ two rows are identical)

Hence, option B is correct.