Câu 1
Ta có
$S_n = \dfrac{1}{1.2} + \dfrac{1}{2.3} + \cdots + \dfrac{1}{n(n+1)}$
$= 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \cdots + \dfrac{1}{n} - \dfrac{n+1}$
$= 1 - \dfrac{n+1} = \dfrac{n}{n+1}$
Ko có đáp án đúng.
Câu 2
Ta có
$S_n = \dfrac{1}{1.3} + \dfrac{1}{3.5} + \cdots + \dfrac{1}{(2n-1)(2n+1)}$
Vậy
$2S_n = \dfrac{2}{1.3} + \dfrac{2}{3.5} + \cdots + \dfrac{2}{(2n-1)(2n+1)}$
$= 1 - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{5} + \cdots + \dfrac{1}{2n-1} - \dfrac{1}{2n+1}$
$= 1 - \dfrac{1}{2n+1} = \dfrac{2n}{2n+1}$
Vậy $S_n = \dfrac{n}{2n+1}$
Đáp án B.
