The correct answer is $\frac{6}{85}$.

To evaluate the value of the given summation series, let us first analyze the structure of each denominator:

$35=5\times 7$

$63=7\times 9$

$99=9\times 11$

$\cdots$

$255=15\times 17$

Notice that every term in the series is of the general form $\frac{1}{n(n+2)}$, where the two factors in the denominator have a common difference of 2.

We can decompose any fraction of this form into partial fractions using the following standard identity:

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

Using this identity, we rewrite each term of the series individually:

$\frac{1}{35}=\frac{1}{5\times 7}=\frac{1}{2}(\frac{1}{5}-\frac{1}{7})$

$\frac{1}{63}=\frac{1}{7\times 9}=\frac{1}{2}(\frac{1}{7}-\frac{1}{9})$

$\frac{1}{99}=\frac{1}{9\times 11}=\frac{1}{2}(\frac{1}{9}-\frac{1}{11})$

$⋮$

$\frac{1}{255}=\frac{1}{15\times 17}=\frac{1}{2}(\frac{1}{15}-\frac{1}{17})$

Now, we sum all the terms together and factor out the common multiplier of $\frac{1}{2}$:

$\text{Sum}=\frac{1}{2}\left[\right(\frac{1}{5}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{9})+(\frac{1}{9}-\frac{1}{11})+\cdots +(\frac{1}{15}-\frac{1}{17}\left)\right]$

This is a telescoping series, which means all intermediate positive and negative terms cancel each other out:

$-\frac{1}{7}+\frac{1}{7}=0$, $-\frac{1}{9}+\frac{1}{9}=0$, $\cdots$, $-\frac{1}{15}+\frac{1}{15}=0$

After these cancellations, only the first term $\frac{1}{5}$ and the last term $\frac{1}{17}$ remain inside the brackets:

$\text{Sum}=\frac{1}{2}(\frac{1}{5}-\frac{1}{17})$

Find the common denominator of $5$ and $17$, which is $85$:

$\frac{1}{5}-\frac{1}{17}=\frac{17-5}{85}=\frac{12}{85}$

Finally, multiply this result by $\frac{1}{2}$:

$\text{Sum}=\frac{1}{2}\times \frac{12}{85}=\frac{6}{85}$

Thus, the final evaluated value of the summation series is $\frac{6}{85}$.