The sum of the series ` 1+1/3*1/4 +1/5*1/(4^(2))+1/7*1/(4^(3)) +...` is

Options

A. `log_(e)1`
B. `log_(e)2`
C. `log_(e)3`
D. `log_(e)4`

Explanation

To find the sum of the series \( S = 1 + \frac{1}{3 \cdot 4} + \frac{1}{5 \cdot 4^2} + \frac{1}{7 \cdot 4^3} + \ldots \), we can approach the problem using the properties of logarithmic functions and their series expansions.

### Step-by-Step Solution:

1. **Identify the Series**:
The series can be expressed as:
\[ S = 1 + \sum_{n=1}^{\infty} \frac{1}{(2n+1) \cdot 4^n} \]

2. **Use Logarithmic Expansion**:
Recall the Taylor series expansion for \(\log(1+x)\):
\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
For \(\log(1-x)\):
\[ \log(1-x) = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots\right) \]

3. **Subtract the Two Expansions**:
If we subtract the expansion of \(\log(1-x)\) from \(\log(1+x)\):
\[ \log(1+x) - \log(1-x) = 2\left(x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots\right) \]
This gives:
\[ \log\left(\frac{1+x}{1-x}\right) = 2\left(x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots\right) \]

4. **Substitute \(x = \frac{1}{2}\)**:
Substitute \(x = \frac{1}{2}\):
\[ \log\left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right) = \log(3) = 2\left(\frac{1}{2} + \frac{1/2^3}{3} + \frac{1/2^5}{5} + \ldots\right) \]
Simplifying gives:
\[ \log(3) = 2\left(\frac{1}{2} + \frac{1}{24} + \frac{1}{160} + \ldots\right) \]

5. **Relate to Our Series**:
The series we have can be related to the series derived from the logarithmic expansion:
\[ S = 1 + \sum_{n=1}^{\infty} \frac{1}{(2n+1) \cdot 4^n} \]
This is equal to \(\log(3)\).

6. **Final Result**:
Thus, the sum of the series is:
\[ S = \log(3) \]

### Conclusion:
The sum of the series \( 1 + \frac{1}{3 \cdot 4} + \frac{1}{5 \cdot 4^2} + \frac{1}{7 \cdot 4^3} + \ldots \) is \( \log(3) \).