If the value of `prod_(k=1)^(50)[{:(1,2k-1),(0,1):}]` is equal to `[{:(1,r),(0,1):}]` then `r` is equal to

Options

A. `62500`
B. `2500`
C. `1250`
D. `12500`

Explanation

To solve the problem, we need to evaluate the product of matrices from \( k = 1 \) to \( k = 50 \) and find the value of \( r \) in the resulting matrix.

The matrices are given in the form:

\[ \begin{pmatrix} 1 & 2k - 1 \\ 0 & 1 \end{pmatrix} \]

### Step 1: Understand the Matrix Multiplication
When we multiply two matrices of the form:

\[ A = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} \]

The product \( AB \) is:

\[ AB = \begin{pmatrix} 1 \cdot 1 + a \cdot 0 & 1 \cdot b + a \cdot 1 \\ 0 \cdot 1 + 1 \cdot 0 & 0 \cdot b + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 1 & a + b \\ 0 & 1 \end{pmatrix} \]

### Step 2: Calculate the Product for k = 1 to 50
For \( k = 1 \) to \( k = 50 \), the matrices are:

\[ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}, \ldots, \begin{pmatrix} 1 & 99 \\ 0 & 1 \end{pmatrix} \]

The second element in each matrix is \( 2k - 1 \) for \( k = 1, 2, \ldots, 50 \).

### Step 3: Sum the Second Elements
When we multiply all these matrices together, the resulting matrix will have the form:

\[ \begin{pmatrix} 1 & S \\ 0 & 1 \end{pmatrix} \]

where \( S \) is the sum of all the second elements:

\[ S = (1) + (3) + (5) + \ldots + (99) \]

This is an arithmetic series where:

- The first term \( a = 1 \)
- The last term \( l = 99 \)
- The number of terms \( n = 50 \)

The sum of an arithmetic series is given by:

\[ S = \frac{n}{2} (a + l) = \frac{50}{2} (1 + 99) = 25 \cdot 100 = 2500 \]

### Step 4: Write the Final Result
Thus, the resulting matrix after multiplying all 50 matrices is:

\[ \begin{pmatrix} 1 & 2500 \\ 0 & 1 \end{pmatrix} \]

This means that \( r = 2500 \).

### Conclusion
The value of \( r \) is:

\[ \boxed{2500} \]