The sum of the series
`1 + ((1)/(2) + (1)/(3)) (1)/(4) + ((1)/(4) + (1)/(5)) (1)/(4^(2)) + ((1)/(6) + (1)/(7)) (1)/(4^(3)) + ...oo`

Options

A. `log sqrt(6)`
B. `log sqrt(3)`
C. `log sqrt(12)`
D. none

Answer

To find the sum of the series

\[ S = 1 + \left( \frac{1}{2} + \frac{1}{3} \right) \frac{1}{4} + \left( \frac{1}{4} + \frac{1}{5} \right) \frac{1}{4^2} + \left( \frac{1}{6} + \frac{1}{7} \right) \frac{1}{4^3} + \ldots \]

we can break it down step by step.

### Step 1: Expand the Series
First, we can rewrite the series by distributing the terms:

\[ S = 1 + \frac{1}{2} \cdot \frac{1}{4} + \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{4} \cdot \frac{1}{4^2} + \frac{1}{5} \cdot \frac{1}{4^2} + \frac{1}{6} \cdot \frac{1}{4^3} + \frac{1}{7} \cdot \frac{1}{4^3} + \ldots \]

### Step 2: Group the Terms
Next, we can group the terms based on the powers of \( \frac{1}{4} \):

\[ S = 1 + \left( \frac{1}{2} + \frac{1}{3} \right) \frac{1}{4} + \left( \frac{1}{4} + \frac{1}{5} \right) \frac{1}{4^2} + \left( \frac{1}{6} + \frac{1}{7} \right) \frac{1}{4^3} + \ldots \]

### Step 3: Identify Patterns
Notice that the terms can be expressed in a more general form. The \( n \)-th term can be represented as:

\[ \left( \frac{1}{2n} + \frac{1}{2n+1} \right) \frac{1}{4^{n-1}} \]

for \( n = 1, 2, 3, \ldots \).

### Step 4: Use Logarithmic Series
Using the logarithmic series, we know:

\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]

By manipulating the series, we can relate it to logarithmic functions.

### Step 5: Substitute Values
We will substitute \( x = \frac{1}{2} \) into the logarithmic series:

\[ S = -\frac{1}{2} \log(1 - x^2) + \frac{1}{2} \log\left(\frac{1+x}{1-x}\right) \]

### Step 6: Simplify
Now substituting \( x = \frac{1}{2} \):

\[ S = -\frac{1}{2} \log\left(1 - \left(\frac{1}{2}\right)^2\right) + \frac{1}{2} \log\left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right) \]

This simplifies to:

\[ S = -\frac{1}{2} \log\left(\frac{3}{4}\right) + \frac{1}{2} \log(3) \]

### Step 7: Final Calculation
Combining the logarithmic terms:

\[ S = \frac{1}{2} \log(4) + \frac{1}{2} \log(3) = \frac{1}{2} \log(12) \]

Thus, the sum of the series is:

\[ S = \frac{1}{2} \log(12) \]