The sum of the first 35 terms of the series `(1)/(2) + (1)/(3) -(1)/(4) -(1)/(2) -(1)/(3) +(1)/(4) +(1)/(2) + (1)/(3) -(1)/(4)`

Options

A. `-(1)/(2)`
B. `-(1)/(4)`
C. `(1)/(4)`
D. 0

Answer

To find the sum of the first 35 terms of the series

\[ \frac{1}{2} + \frac{1}{3} - \frac{1}{4} - \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \]

we can observe the pattern in the series.

### Step 1: Identify the repeating pattern
The series can be grouped into sets of 6 terms:
1. \( \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \)
2. \( -\frac{1}{2} - \frac{1}{3} + \frac{1}{4} \)
3. \( +\frac{1}{2} + \frac{1}{3} - \frac{1}{4} \)

This pattern repeats every 6 terms.

### Step 2: Calculate the sum of one complete cycle (6 terms)
Let's calculate the sum of these 6 terms:
\[ \left(\frac{1}{2} + \frac{1}{3} - \frac{1}{4}\right) + \left(-\frac{1}{2} - \frac{1}{3} + \frac{1}{4}\right) \]

Calculating the first group:
\[ \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{6}{12} + \frac{4}{12} - \frac{3}{12} = \frac{7}{12} \]

Calculating the second group:
\[ -\frac{1}{2} - \frac{1}{3} + \frac{1}{4} = -\frac{6}{12} - \frac{4}{12} + \frac{3}{12} = -\frac{7}{12} \]

Adding these two results:
\[ \frac{7}{12} - \frac{7}{12} = 0 \]

Thus, the sum of every complete cycle of 6 terms is 0.

### Step 3: Determine how many complete cycles fit into 35 terms
To find how many complete cycles of 6 fit into 35, we perform the division:
\[ 35 \div 6 = 5 \quad \text{(with a remainder of 5)} \]

This means there are 5 complete cycles (which contribute 0 to the sum) and 5 additional terms.

### Step 4: Identify the remaining 5 terms
The remaining 5 terms after 30 terms (5 complete cycles) are:
1. \( \frac{1}{2} \)
2. \( \frac{1}{3} \)
3. \( -\frac{1}{4} \)
4. \( -\frac{1}{2} \)
5. \( -\frac{1}{3} \)

### Step 5: Calculate the sum of the remaining 5 terms
Now, we sum these remaining terms:
\[ \frac{1}{2} + \frac{1}{3} - \frac{1}{4} - \frac{1}{2} - \frac{1}{3} \]

The \( \frac{1}{2} \) and \( -\frac{1}{2} \) cancel out, and the \( \frac{1}{3} \) and \( -\frac{1}{3} \) also cancel out. Thus, we are left with:
\[ -\frac{1}{4} \]

### Final Answer
The sum of the first 35 terms of the series is:
\[ \boxed{-\frac{1}{4}} \]