The value of

$\frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + ... + \frac{1}{27 \times 31}$ is:

The given series is:

$\frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + ... + \frac{1}{27 \times 31}$

The numbers follow the pattern: 3, 7, 11, 15, ..., 27, 31, where each number increases by 4.

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Using partial fractions, each term can be decomposed as:

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

Since consecutive terms differ by 4:

$\frac{1}{3 \times 7} = \frac{1}{4}\left(\frac{1}{3} - \frac{1}{7}\right)$

$\frac{1}{7 \times 11} = \frac{1}{4}\left(\frac{1}{7} - \frac{1}{11}\right)$

$\frac{1}{11 \times 15} = \frac{1}{4}\left(\frac{1}{11} - \frac{1}{15}\right)$

$\vdots$

$\frac{1}{27 \times 31} = \frac{1}{4}\left(\frac{1}{27} - \frac{1}{31}\right)$

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Adding all terms:

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

This is a telescoping series where intermediate terms cancel:

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

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

$= \frac{1}{4} \times \frac{28}{93}$

$= \frac{28}{372}$

$= \frac{7}{93}$

Therefore, the value is $\frac{7}{93}$.