The sum of series ` 2[ 7^(-1)+3^(-1).7^(-3)+5^(-1).7^(-5)+...]` is

Options

A. `log_(e)(4/3)`
B. `log_(e)(3/4)`
C. `2log_(e)(3/4)`
D. `2log_(e)(4/3)`

Explanation

To find the sum of the series \( S = 2\left( 7^{-1} + 3^{-1} \cdot 7^{-3} + 5^{-1} \cdot 7^{-5} + \ldots \right) \), we can follow these steps:

### Step 1: Rewrite the series
The series can be rewritten as:
\[ S = 2\left( \frac{1}{7} + \frac{1}{3 \cdot 7^3} + \frac{1}{5 \cdot 7^5} + \ldots \right) \]

### Step 2: Identify the general term
The general term of the series can be expressed as:
\[ \frac{1}{(2n-1) \cdot 7^{2n-1}} \quad \text{for } n = 1, 2, 3, \ldots \]
Thus, we can write:
\[ S = 2 \sum_{n=1}^{\infty} \frac{1}{(2n-1) \cdot 7^{2n-1}} \]

### Step 3: Recognize the series
This series resembles the Taylor series expansion for the logarithm. We know that:
\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
and
\[ \log(1-x) = -\left( x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots \right) \]

### Step 4: Use the logarithmic identity
The difference between the logarithms gives us:
\[ \log(1+x) - \log(1-x) = 2\left( x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots \right) \]
So we can express our series as:
\[ \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2n-1} = \frac{1}{2} \left( \log(1+x) - \log(1-x) \right) \]

### Step 5: Substitute \( x = \frac{1}{7} \)
Substituting \( x = \frac{1}{7} \):
\[ \log\left(1 + \frac{1}{7}\right) - \log\left(1 - \frac{1}{7}\right) = \log\left(\frac{8}{6}\right) = \log\left(\frac{4}{3}\right) \]

### Step 6: Calculate the sum
Thus, we have:
\[ S = 2 \cdot \frac{1}{2} \log\left(\frac{4}{3}\right) = \log\left(\frac{4}{3}\right) \]

### Final Answer
The sum of the series is:
\[ \boxed{\log\left(\frac{4}{3}\right)} \]