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 find the limit as \( n \) approaches infinity of the summation of the determinants of the matrix \( A_r \) from \( r = 1 \) to \( n \).

### Step-by-Step Solution:

1. **Define the Matrix \( A_r \)**:
\[ A_r = \begin{pmatrix} \frac{1}{r(r+1)} & \frac{1}{3^r} \\ 2 & 3 \end{pmatrix} \]

2. **Calculate the Determinant of \( A_r \)**:
The determinant of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by \( ad - bc \). Therefore, for our matrix:
\[ |A_r| = \left(\frac{1}{r(r+1)} \cdot 3\right) - \left(2 \cdot \frac{1}{3^r}\right) \]
Simplifying this gives:
\[ |A_r| = \frac{3}{r(r+1)} - \frac{2}{3^r} \]

3. **Set Up the Summation**:
We need to find:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} |A_r| = \lim_{n \to \infty} \sum_{r=1}^{n} \left(\frac{3}{r(r+1)} - \frac{2}{3^r}\right) \]

4. **Separate the Summation**:
This can be separated into two summations:
\[ \lim_{n \to \infty} \left( \sum_{r=1}^{n} \frac{3}{r(r+1)} - \sum_{r=1}^{n} \frac{2}{3^r} \right) \]

5. **Evaluate the First Summation**:
The first summation can be simplified using the identity:
\[ \frac{3}{r(r+1)} = 3\left(\frac{1}{r} - \frac{1}{r+1}\right) \]
This is a telescoping series:
\[ \sum_{r=1}^{n} \frac{3}{r(r+1)} = 3\left(1 - \frac{1}{n+1}\right) = \frac{3n}{n+1} \]

6. **Evaluate the Second Summation**:
The second summation is a geometric series:
\[ \sum_{r=1}^{n} \frac{2}{3^r} = 2 \left(\frac{1/3}{1 - 1/3}\right) = 2 \cdot \frac{1/3}{2/3} = 1 - \frac{2}{3^n} \]

7. **Combine the Results**:
Now we combine the results of both summations:
\[ \lim_{n \to \infty} \left( \frac{3n}{n+1} - \left(1 - \frac{2}{3^n}\right) \right) \]
As \( n \to \infty \), \( \frac{3n}{n+1} \to 3 \) and \( \frac{2}{3^n} \to 0 \):
\[ = 3 - 1 = 2 \]

8. **Final Result**:
Therefore, the limit is:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} |A_r| = 2 \]