A) \[2013\]
B) \[2014\]
C) \[{{(2013)}^{2}}\]
D) \[{{(2014)}^{2}}\]
Correct Answer: D
Solution :
[d] det \[(Mr)=\left| \begin{matrix} r & r-1 \\ r-1 & r \\ \end{matrix} \right|=2r-1\] \[\sum\limits_{r=1}^{2014}{\det ({{M}_{r}})=2\sum\limits_{r=1}^{2014}{r-2014}}\] \[=2\times \frac{2014\times 2015}{2}-2014={{(2014)}^{2}}\]You need to login to perform this action.
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