Chọn phương án A.

\(\begin{eqnarray*}
S_n&=&\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\cdots+\dfrac{1}{n\cdot(n+1)}\\
&=&\dfrac{2-1}{1\cdot2}+\dfrac{3-2}{2\cdot3}+\dfrac{4-3}{3\cdot4}+\cdots+\dfrac{(n+1)-n}{n\cdot(n+1)}\\
&=&\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots+\dfrac{1}{n}-\dfrac{1}{n+1}\\
&=&\dfrac{1}{1}-\dfrac{1}{n+1}\\
&=&\dfrac{n}{n+1}
\end{eqnarray*}\)

Vậy \(S_{2017}=\dfrac{2017}{2018}\).

Chọn phương án A.

Ta có:

• \(S_1=\dfrac{1}{2}\)
• \(S_2=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}=\dfrac{2}{3}\)
• \(S_3=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}=\dfrac{3}{4}\)

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

Vậy \(S_{2017}=\dfrac{2017}{2018}\).