Cho tổng \(S_n=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\cdots+\dfrac{1}{(2n-1)(2n+1)}\) với \(n\in\Bbb{N}^*\). Tính \(S_{2017}\).
![]() | \(S_{2017}=\dfrac{2017}{2018}\) |
![]() | \(S_{2017}=\dfrac{2017}{4035}\) |
![]() | \(S_{2017}=\dfrac{1}{2018}\) |
![]() | \(S_{2017}=\dfrac{2017}{4033}\) |
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:
Theo đó, \(S_n=\dfrac{n}{2n+1}\) với \(n\in\Bbb{N}^*\).
Vậy \(S_{2017}=\dfrac{2017}{4035}\).