If `A= (1)/(1 xx 2) + (1)/(1 xx 4) + (1)/(2 xx 3) + (1)/(4 xx 7) + (1)/(3 xx 4) + (1)/(7 xx 10)` …..upto 20 terms, then what is the value of A?

Options

A. 379/308
B. 171/140
C. 379/310
D. 420/341

Answer

To solve the problem, we need to evaluate the sum \( A \) given by:

\[ A = \frac{1}{1 \times 2} + \frac{1}{1 \times 4} + \frac{1}{2 \times 3} + \frac{1}{4 \times 7} + \frac{1}{3 \times 4} + \frac{1}{7 \times 10} + \ldots \text{ (up to 20 terms)} \]

### Step 1: Identify Patterns in the Series
The series consists of two distinct patterns:
1. The first pattern is of the form \( \frac{1}{n(n+1)} \).
2. The second pattern appears to be of the form \( \frac{1}{(3n-2)(3n+1)} \).

### Step 2: Break Down the Series
We can separate the series into two parts:
- The first part includes terms of the form \( \frac{1}{n(n+1)} \).
- The second part includes terms of the form \( \frac{1}{(3n-2)(3n+1)} \).

### Step 3: Calculate the First Pattern
For the first pattern, we have:
\[ \sum_{n=1}^{10} \frac{1}{n(n+1)} = \sum_{n=1}^{10} \left( \frac{1}{n} - \frac{1}{n+1} \right) \]
This is a telescoping series, which simplifies to:
\[ 1 - \frac{1}{11} = \frac{10}{11} \]

### Step 4: Calculate the Second Pattern
For the second pattern, we have:
\[ \sum_{n=1}^{10} \frac{1}{(3n-2)(3n+1)} \]
Using partial fractions, we can express:
\[ \frac{1}{(3n-2)(3n+1)} = \frac{A}{3n-2} + \frac{B}{3n+1} \]
Solving for \( A \) and \( B \), we find:
\[ A = \frac{1}{3}, \quad B = -\frac{1}{3} \]
Thus, we can rewrite the sum as:
\[ \sum_{n=1}^{10} \left( \frac{1}{3(3n-2)} - \frac{1}{3(3n+1)} \right) \]
This also forms a telescoping series:
\[ \frac{1}{3} \left( \frac{1}{1} - \frac{1}{31} \right) = \frac{1}{3} \left( 1 - \frac{1}{31} \right) = \frac{1}{3} \cdot \frac{30}{31} = \frac{10}{31} \]

### Step 5: Combine the Results
Now, we can combine both parts:
\[ A = \frac{10}{11} + \frac{10}{31} \]
To add these fractions, we find a common denominator:
\[ \text{LCM of } 11 \text{ and } 31 = 341 \]
Thus,
\[ A = \frac{10 \times 31}{341} + \frac{10 \times 11}{341} = \frac{310 + 110}{341} = \frac{420}{341} \]

### Final Answer
The value of \( A \) is:
\[ \boxed{\frac{420}{341}} \]