Chọn phương án B.

\(\begin{eqnarray*}
S_n&=&\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\cdots+\dfrac{1}{(2n-1)(2n+1)}\\
&=&\dfrac{1}{2}\left(\dfrac{3-1}{1\cdot3}+\dfrac{5-3}{3\cdot5}+\dfrac{7-5}{5\cdot7}+\cdots+\dfrac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}\right)\\
&=&\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\cdots+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\
&=&\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{2n+1}\right)\\
&=&\dfrac{n}{2n+1}.
\end{eqnarray*}\)

Vậy \(S_{2017}=\dfrac{2017}{4035}\).

Chọn phương án B.

Ta có:

• \(S_1=\dfrac{1}{1\cdot3}=\dfrac{1}{3}=\dfrac{1}{2\cdot1-1}\)
• \(S_2=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}=\dfrac{2}{5}=\dfrac{1}{2\cdot2+1}\)
• \(S_3=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}=\dfrac{3}{7}=\dfrac{1}{2\cdot3+1}\)

Theo đó, \(S_n=\dfrac{n}{2n+1}\) với \(n\in\Bbb{N}^*\).

Vậy \(S_{2017}=\dfrac{2017}{4035}\).