$$\Sigma\ _{n=1}^{15}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+2\right)\left(n+3\right)}\right)=$$

The value of the given telescoping series is $$\frac{95}{144}$$.

To evaluate the sum, rewrite each partial fraction using the standard decomposition:

$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$

$$\frac{1}{(n+2)(n+3)} = \frac{1}{n+2} - \frac{1}{n+3}$$

Substitute these into the summation:

$$\sum_{n=1}^{15} \left[ \left(\frac{1}{n} - \frac{1}{n+1}\right) - \left(\frac{1}{n+2} - \frac{1}{n+3}\right) \right]$$

This can be split into two separate telescoping series:

$$\sum_{n=1}^{15} \left(\frac{1}{n} - \frac{1}{n+1}\right) - \sum_{n=1}^{15} \left(\frac{1}{n+2} - \frac{1}{n+3}\right)$$

Evaluate the first sum:

$$\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{15} - \frac{1}{16}\right) = 1 - \frac{1}{16}$$

Evaluate the second sum (shifting indices or expanding):

$$\left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \dots + \left(\frac{1}{17} - \frac{1}{18}\right) = \frac{1}{3} - \frac{1}{18}$$

Subtract the two evaluated sums:

$$\left(1 - \frac{1}{16}\right) - \left(\frac{1}{3} - \frac{1}{18}\right) = 1 - \frac{1}{3} - \frac{1}{16} + \frac{1}{18}$$

$$= \frac{2}{3} - \left(\frac{9 - 8}{144}\right)$$

$$= \frac{2}{3} - \frac{1}{144}$$

$$= \frac{96 - 1}{144} = \frac{95}{144}$$

The correct option is B.