If `A_(i)= [(2^(-i),3^(-i)),(3^(-i),2^(-i))],then sum_(i=1)^(oo) det (A_(i))` is equal to

Options

A. `(3)/(4)`
B. `(5)/(24)`
C. `(5)/(4)`
D. `(7)/(144)`

Answer

To solve the problem, we need to find the sum of the determinants of the matrices \( A_i \) defined as:

\[ A_i = \begin{pmatrix} 2^{-i} & 3^{-i} \\ 3^{-i} & 2^{-i} \end{pmatrix} \]

We will calculate the determinant of \( A_i \) and then find the infinite sum of these determinants.

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

The determinant of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by \( ad - bc \).

For our matrix \( A_i \):

\[ \text{det}(A_i) = (2^{-i})(2^{-i}) - (3^{-i})(3^{-i}) = 2^{-2i} - 3^{-2i} \]

### Step 2: Set up the infinite sum

Now we need to find the sum:

\[ \sum_{i=1}^{\infty} \text{det}(A_i) = \sum_{i=1}^{\infty} (2^{-2i} - 3^{-2i}) \]

This can be separated into two sums:

\[ \sum_{i=1}^{\infty} 2^{-2i} - \sum_{i=1}^{\infty} 3^{-2i} \]

### Step 3: Evaluate the first sum

The first sum is a geometric series:

\[ \sum_{i=1}^{\infty} 2^{-2i} = \sum_{i=1}^{\infty} \left( \frac{1}{4} \right)^i \]

The formula for the sum of an infinite geometric series \( \sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r} \) applies here, where \( a = \frac{1}{4} \) and \( r = \frac{1}{4} \):

\[ \sum_{i=1}^{\infty} 2^{-2i} = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \]

### Step 4: Evaluate the second sum

Now for the second sum:

\[ \sum_{i=1}^{\infty} 3^{-2i} = \sum_{i=1}^{\infty} \left( \frac{1}{9} \right)^i \]

Using the same formula for the geometric series, where \( a = \frac{1}{9} \) and \( r = \frac{1}{9} \):

\[ \sum_{i=1}^{\infty} 3^{-2i} = \frac{\frac{1}{9}}{1 - \frac{1}{9}} = \frac{\frac{1}{9}}{\frac{8}{9}} = \frac{1}{8} \]

### Step 5: Combine the results

Now we combine the results of the two sums:

\[ \sum_{i=1}^{\infty} \text{det}(A_i) = \frac{1}{3} - \frac{1}{8} \]

To subtract these fractions, we need a common denominator. The least common multiple of 3 and 8 is 24:

\[ \frac{1}{3} = \frac{8}{24}, \quad \frac{1}{8} = \frac{3}{24} \]

Thus,

\[ \frac{1}{3} - \frac{1}{8} = \frac{8}{24} - \frac{3}{24} = \frac{5}{24} \]

### Final Answer

The value of the infinite sum is:

\[ \sum_{i=1}^{\infty} \text{det}(A_i) = \frac{5}{24} \]

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