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If the matrix \(M_r\) is given by \(M_r=\left(\begin{array}{cc}r & r-1 \\ r-1 & r\end{array}\right)\) for \(r=1,2,3, \ldots\) then \(\operatorname{det}\left(\mathrm{M}_1\right)+\operatorname{det}\left(\mathrm{M}_2\right)+\ldots . .+\operatorname{det}\left(\mathrm{M}_{2008}\right)=\)
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Verified Answer
The correct answer is:
\((2008)^2\)
\(\begin{aligned}
& \text {Hint : } \operatorname{det}\left(M_r\right)=r^2-(r-1)^2=r^2-\left(r^2-2 r+1\right)=2 r-1 \\
& \begin{aligned}
\therefore \sum_{r=1}^{2008} \operatorname{det}\left(M_r\right)=\sum_{r=1}^{2008}(2 r-1)= & +3+5+\cdots+4015 \\
= & (2008)^2
\end{aligned}
\end{aligned}\)
& \text {Hint : } \operatorname{det}\left(M_r\right)=r^2-(r-1)^2=r^2-\left(r^2-2 r+1\right)=2 r-1 \\
& \begin{aligned}
\therefore \sum_{r=1}^{2008} \operatorname{det}\left(M_r\right)=\sum_{r=1}^{2008}(2 r-1)= & +3+5+\cdots+4015 \\
= & (2008)^2
\end{aligned}
\end{aligned}\)
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