value of det \[({{M}_{1}})+\det ({{M}_{2}})+\det ({{M}_{3}})+....+\det ({{M}_{2007}})\] is
A) 2007
B) 2008
C) 20082
D) 20072
Correct Answer: D
Solution :
det\[({{M}_{r}})=\left| \begin{matrix} r & r-1 \\ r-1 & r \\ \end{matrix} \right|=2r-1\] \[\sum\limits_{r=l}^{2007}{\det }({{M}_{r}})=2\sum\limits_{r=l}^{2007}{r-2007}\] \[=2\times \frac{2007\times 2008}{2}-2007={{(2007)}^{2}}\]
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