Matrix `M_(r)` is defined as `M_(r) = ({:(r,r-1),(r-1,r):}), r in ` N. The value of det `(M_(1)) + det (M_(2)) + det (M_(3))+ .... ) det (M_(2014))` is

Options

A. 2013
B. 2014
C. `(2013)^(2)`
D. `(2014)^(2)`

Answer

To solve the problem, we need to calculate the value of the sum of determinants of the matrices \( M_r \) defined as:

\[ M_r = \begin{pmatrix} r & r-1 \\ r-1 & r \end{pmatrix} \]

We need to find:

\[ \text{det}(M_1) + \text{det}(M_2) + \text{det}(M_3) + \ldots + \text{det}(M_{2014}) \]

### Step 1: Calculate the determinant of \( M_r \)

The determinant of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by \( ad - bc \). For our matrix \( M_r \):

\[ \text{det}(M_r) = r \cdot r - (r-1)(r-1) = r^2 - (r^2 - 2r + 1) = 2r - 1 \]

### Step 2: Set up the summation

Now we can express the desired sum as:

\[ \sum_{r=1}^{2014} \text{det}(M_r) = \sum_{r=1}^{2014} (2r - 1) \]

### Step 3: Simplify the summation

We can separate the summation:

\[ \sum_{r=1}^{2014} (2r - 1) = \sum_{r=1}^{2014} 2r - \sum_{r=1}^{2014} 1 \]

### Step 4: Calculate each part of the summation

1. **Calculate \( \sum_{r=1}^{2014} 2r \)**:
\[ \sum_{r=1}^{2014} 2r = 2 \sum_{r=1}^{2014} r = 2 \cdot \frac{2014 \cdot 2015}{2} = 2014 \cdot 2015 \]

2. **Calculate \( \sum_{r=1}^{2014} 1 \)**:
\[ \sum_{r=1}^{2014} 1 = 2014 \]

### Step 5: Combine the results

Now substituting back into our equation:

\[ \sum_{r=1}^{2014} (2r - 1) = 2014 \cdot 2015 - 2014 = 2014 (2015 - 1) = 2014 \cdot 2014 \]

### Final Answer

Thus, the value of \( \text{det}(M_1) + \text{det}(M_2) + \text{det}(M_3) + \ldots + \text{det}(M_{2014}) \) is:

\[ \boxed{2014^2} \]