If $$ A_{r}=\left|\begin{array}{cc} r & r-1 \\ r-1 & r \end{array}\right| $$ where \(\mathrm{r}\) is a natural number than the value of \(\left.\sqrt{[}^{2013} \sum_{r=1} A_{r}\right]\) is (a) 1 (b) 40 (c) 2012 (d) 2013

To compute the determinant A_r of a 2x2 matrix, we will find the difference between the product of the diagonal elements (top-left and bottom-right) and the product of the off-diagonal elements (top-right and bottom-left). Therefore, we have: A_r = |A| = \[ (r \cdot r) - ((r-1) \cdot (r-1)) \]

Now, we need to calculate the sum of the determinants for all values of r from 1 to 2013: \[ S = \sum_{r=1}^{2013} A_r \] We will replace A_r with the determinant expression we obtained in Step 1 and perform the summation: \[ S = \sum_{r=1}^{2013} (r^2 - (r-1)^2) \]

To simplify the sum expression, let's expand the square terms and group the terms: \[ S = \sum_{r=1}^{2013} (r^2 - r^2 + 2r - 1) \] Now, cancel out the r^2 terms and simplify the remaining expression: \[ S = \sum_{r=1}^{2013} (2r - 1) \] Now sum up the remaining terms: \[ S = (2\cdot1 - 1) + (2\cdot2 - 1) + \cdots + (2\cdot2013 - 1) \] Since this is an arithmetic series with common difference 2, we can use the formula to calculate the sum: \[ S = \frac{n}{2} (2a + (n-1)d) \] where n is the number of terms, a is the first term, and d is the common difference: \[ S = \frac{2013}{2} (2(2\cdot1-1) + (2013-1)(2)) = 2013\cdot2012 \] Now we need to take the square root of the sum: \[ \sqrt{S} = \sqrt{2013\cdot2012} \]

Finally, let's compare the square root of the sum with the given options: (a) 1 (b) 40 (c) 2012 (d) 2013 Since \[\sqrt{2013\cdot2012} \approx 2012 \] is close to option (c), we will choose: Answer: (c) 2012