Question

Let $m$ be a positive integer and $\Delta_{r}=\begin{vmatrix} 2 r-1 & { }^{m} C_{r} & 1 \\ m^{2}-1 & 2^{m} & m+1 \\ \sin ^{2}\left(m^{2}\right) & \sin ^{2}(m) & \sin ^{2}(m+1) \end{vmatrix}(0 \leq r \leq m)$ Then the value of $\displaystyle\sum_{r=0}^{m} \Delta_{r}$ 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 $\displaystyle\sum_{r=0}^{m} \Delta_{r}$
$=\begin{vmatrix}\displaystyle\sum_{r=0}^{m}(2 r-1) & \displaystyle\sum_{r=0}^{m} C_{r} & \displaystyle\sum_{r=0}^{m} 1 \\m^{2}-1 & 2^{m} & m+1 \\\sin ^{2}\left(m^{2}\right) & \sin ^{2}(m) & \sin ^{2}(m+1)\end{vmatrix}$
But $\displaystyle\sum_{r=0}^{m}(2 r-1)=\frac{1}{2}(m+1)(2 m-1-1)=m^{2}-1$,
$\displaystyle\sum_{r=0}^{m}{ }^{m} C_{r}=2^{m}$ and $\displaystyle\sum_{r=0}^{m} 1=m+1$
Therefore, $\displaystyle\sum_{r=0}^{m} \Delta_{r}=\begin{vmatrix}m^{2}-1 & 2^{m} & m+1 \\ m^{2}-1 & 2^{m} & m+1 \\ \sin ^{2}\left(m^{2}\right) & \sin ^{2}(m) & \sin ^{2}(m+1)\end{vmatrix}=0$