The value of $\frac{1}{3×7}+\frac{1}{7×11}+\frac{1}{11×15}+....+\frac{1}{27×31}$ is:

$\frac{7}{93}$

The correct answer is Option (2) → $\frac{7}{93}$

1. Analyze the General Term

Each term in the series is of the form $\frac{1}{n \times (n+d)}$, where $d$ is the common difference between the two numbers in the denominator.

In this series:

• First term: $\frac{1}{3 \times 7}$ (Difference $d = 7 - 3 = 4$)
• Second term: $\frac{1}{7 \times 11}$ (Difference $d = 11 - 7 = 4$)
• Last term: $\frac{1}{27 \times 31}$ (Difference $d = 31 - 27 = 4$)

The common difference throughout the series is $4$.

2. Apply the Telescoping Sum Formula

Any term $\frac{1}{a \times b}$ can be written as:

$\frac{1}{a \times b} = \frac{1}{b - a} \left( \frac{1}{a} - \frac{1}{b} \right)$

Applying this to our series:

$S = \frac{1}{4} \left( \frac{1}{3} - \frac{1}{7} \right) + \frac{1}{4} \left( \frac{1}{7} - \frac{1}{11} \right) + \frac{1}{4} \left( \frac{1}{11} - \frac{1}{15} \right) + \dots + \frac{1}{4} \left( \frac{1}{27} - \frac{1}{31} \right)$

3. Simplify the Expression

We can factor out $\frac{1}{4}$:

$S = \frac{1}{4} \left[ \left( \frac{1}{3} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{11} \right) + \left( \frac{1}{11} - \frac{1}{15} \right) + \dots + \left( \frac{1}{27} - \frac{1}{31} \right) \right]$

In a telescoping series, the intermediate terms cancel each other out ($-\frac{1}{7} + \frac{1}{7} = 0$, $-\frac{1}{11} + \frac{1}{11} = 0$, etc.), leaving only the first and the last terms:

$S = \frac{1}{4} \left( \frac{1}{3} - \frac{1}{31} \right)$

4. Final Calculation

$S = \frac{1}{4} \left( \frac{31 - 3}{3 \times 31} \right)$

$S = \frac{1}{4} \left( \frac{28}{93} \right)$

$S = \frac{7}{93}$