Let `S=1/(1.4)+1/(4.7)+1/(7.10)+.....+ n ` terms observe the following lists

Options

A. `(1)/(6)`
B. `(1)/(2)`
C. `(1)/(4)`
D. `(1)/(3)`

Explanation

To solve the problem, we need to find the sum \( S \) of the series given by:

\[ S = \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \ldots \text{ (n terms)} \]

### Step 1: Identify the general term
The denominators of the terms in the series can be expressed as follows:
- The first term is \( 1 \cdot 4 \)
- The second term is \( 4 \cdot 7 \)
- The third term is \( 7 \cdot 10 \)

We can see that the denominators follow a pattern. The \( n \)-th term can be expressed as:
\[ T_n = \frac{1}{(3n - 2)(3n + 1)} \]

### Step 2: Rewrite the sum
Thus, we can rewrite the sum \( S \) as:
\[ S = \sum_{r=1}^{n} \frac{1}{(3r - 2)(3r + 1)} \]

### Step 3: Use partial fractions
Next, we can use partial fractions to simplify the term:
\[ \frac{1}{(3r - 2)(3r + 1)} = \frac{A}{3r - 2} + \frac{B}{3r + 1} \]
Multiplying through by the denominator \((3r - 2)(3r + 1)\) gives:
\[ 1 = A(3r + 1) + B(3r - 2) \]

### Step 4: Solve for A and B
Expanding and combining like terms:
\[ 1 = (3A + 3B)r + (A - 2B) \]

Setting the coefficients equal gives us the system of equations:
1. \( 3A + 3B = 0 \)
2. \( A - 2B = 1 \)

From the first equation, we have \( A + B = 0 \) or \( A = -B \). Substituting into the second equation:
\[ -B - 2B = 1 \implies -3B = 1 \implies B = -\frac{1}{3} \]
Thus, \( A = \frac{1}{3} \).

### Step 5: Rewrite the partial fraction
Now we can rewrite the term:
\[ \frac{1}{(3r - 2)(3r + 1)} = \frac{1/3}{3r - 2} - \frac{1/3}{3r + 1} \]

### Step 6: Substitute back into the sum
Substituting back into the sum gives:
\[ S = \sum_{r=1}^{n} \left( \frac{1/3}{3r - 2} - \frac{1/3}{3r + 1} \right) \]

### Step 7: Simplify the series
This is a telescoping series. When we write out the first few terms, we see that many terms will cancel:
\[ S = \frac{1}{3} \left( \left(\frac{1}{1} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{10}\right) + \ldots \right) \]

### Step 8: Evaluate the limit
After cancellation, we are left with:
\[ S = \frac{1}{3} \left( 1 - \frac{1}{3n + 1} \right) \]
Taking the limit as \( n \) approaches infinity:
\[ \lim_{n \to \infty} S = \frac{1}{3} \left( 1 - 0 \right) = \frac{1}{3} \]

### Final Answer
Thus, the value of \( S \) as \( n \) approaches infinity is:
\[ \boxed{\frac{1}{3}} \]