If `A_(r)=[((1)/(r(r+1)),(1)/(3^(r ))),(2,3)]`, then `lim_(nrarroo)Sigma_(r=1)^(n)|A_(r)|` is equal to

Options

A. 0
B. 2
C. 4
D. 8

Answer

To solve the problem, we need to evaluate the limit as \( n \) approaches infinity of the sum of the determinants of the matrix \( A_r \). The matrix is given as:

\[ A_r = \begin{pmatrix} \frac{1}{r(r+1)} & \frac{1}{3^r} \\ 2 & 3 \end{pmatrix} \]

### Step 1: Calculate the Determinant of \( A_r \)

The determinant of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is calculated as \( ad - bc \). Therefore, for our matrix \( A_r \):

\[ |A_r| = \left(\frac{1}{r(r+1)} \cdot 3\right) - \left(\frac{1}{3^r} \cdot 2\right) \]

Calculating this gives:

\[ |A_r| = \frac{3}{r(r+1)} - \frac{2}{3^r} \]

### Step 2: Set Up the Limit Expression

We need to find:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} |A_r| = \lim_{n \to \infty} \sum_{r=1}^{n} \left( \frac{3}{r(r+1)} - \frac{2}{3^r} \right) \]

This can be separated into two sums:

\[ \lim_{n \to \infty} \left( \sum_{r=1}^{n} \frac{3}{r(r+1)} - \sum_{r=1}^{n} \frac{2}{3^r} \right) \]

### Step 3: Evaluate the First Sum

The first sum can be simplified:

\[ \sum_{r=1}^{n} \frac{3}{r(r+1)} = 3 \sum_{r=1}^{n} \left( \frac{1}{r} - \frac{1}{r+1} \right) \]

This is a telescoping series, which simplifies to:

\[ 3 \left( 1 - \frac{1}{n+1} \right) = 3 - \frac{3}{n+1} \]

Taking the limit as \( n \to \infty \):

\[ \lim_{n \to \infty} \left( 3 - \frac{3}{n+1} \right) = 3 \]

### Step 4: Evaluate the Second Sum

The second sum is a geometric series:

\[ \sum_{r=1}^{n} \frac{2}{3^r} = 2 \left( \frac{1/3}{1 - 1/3} \right) = 2 \left( \frac{1/3}{2/3} \right) = 1 \]

As \( n \to \infty \), this sum converges to 1.

### Step 5: Combine the Results

Now we combine the results from both sums:

\[ \lim_{n \to \infty} \left( 3 - 1 \right) = 2 \]

### Final Answer

Thus, the final result is:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} |A_r| = 2 \]