Question

If $\Delta_r = \begin{vmatrix}2r-1&^{m}C_{r}&1\\ m^{2} - 1 &2^{m}&m+1\\ \sin^{2}\left(m^{2}\right)&\sin^{2} \left(m\right)&\sin^{2} \left(m + 1\right)\end{vmatrix}$ , then the value of $ \displaystyle\sum^{m}_{r=0} \Delta r $, is

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

Answer 1

Answer: (B)

$\Delta_{r}=\begin{vmatrix}2 r-1 & { }^{m} C_{r} & 1 \\ m^{2}-1 & 2^{m} & m+1 \\ \sin ^{2}\left(m^{2}\right) & \sin ^{2}(m) & \sin ^{2}(m+1)\end{vmatrix}$
$\therefore \displaystyle\sum_{r=0}^{m} \Delta_{r}=\begin{vmatrix}\displaystyle\sum_{r=0}^{m}(2 r-1) & \displaystyle\sum_{r=0}^{m}{ }^{m} C_{r} & \displaystyle\sum_{r=0}^{m} 1 \\ m^{2}-1 & 2^{m} & m+1 \\ \sin ^{2}\left(m^{2}\right) & \sin ^{2}(m) & \sin ^{2}(m+1)\end{vmatrix}$
$=\begin{vmatrix}m^{2}-1 & 2^{m} & m+1 \\ m^{2}-1 & 2^{m} & m+1 \\ \sin ^{2}\left(m^{2}\right) & \sin ^{2}(m) & \sin ^{2}(m+1)\end{vmatrix}$
$=0 (\because$ two rows are identical)