Discussion Forum

If $\triangle_{r}=\begin{bmatrix}2r-1 & ^{m}C_{r}&1 \\m^{2}-1 &2^{m}&m+1\\\sin^{2}(m^{2})&\sin^{2}(m) &\sin^{2}(m+1) \end{bmatrix}$, then  the value of  $\sum_{r=0}^{m}\triangle_{r}$  is 

Answer:

Explanation:

$\triangle_{r}=\begin{bmatrix}2r-1 & ^{m}C_{r}&1 \\m^{2}-1 &2^{m}&m+1\\\sin^{2}(m^{2})&\sin^{2}(m) &\sin^{2}(m+1) \end{bmatrix}$

$\therefore$   $\sum_{r=0}^{m}\triangle_{r}=$   $\begin{bmatrix}\sum_{r=0}^{m}{(2r-1)} & \sum_{r=0}^{m} {^{m}}C_{r}&\sum_{r=0}^{m} 1 \\m^{2}-1 &2^{m}&m+1\\\sin^{2}(m^{2})&\sin^{2}(m) &\sin^{2}(m+1) \end{bmatrix}$

 $==\begin{bmatrix}m^{2}-1 & 2^{m}&m+1 \\m^{2}-1 &2^{m}&m+1\\\sin^{2}(m^{2})&\sin^{2}(m) &\sin^{2}(m+1) \end{bmatrix}$

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