Chọn câu đúng.

Xét đáp án A:

$\begin{array}{l}S = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + ... + \dfrac{1}{{{{50}^2}}}\\ < \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + .... + \dfrac{1}{{49.50}}\\ = 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + .... + \dfrac{1}{{49}} - \dfrac{1}{{50}}\\ = 1 - \dfrac{1}{{50}} = \dfrac{{49}}{{50}} < 1\end{array}$

Vậy $S < 1$.

Vậy A sai.

Xét đáp án B: $S = \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{49.50}}$

$\begin{array}{l} S = 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... - \dfrac{1}{{49}} + \dfrac{1}{{49}} - \dfrac{1}{{50}}\\ = 1 - \dfrac{1}{{50}} = \dfrac{{49}}{{50}} < 1 \end{array}$

Vậy B sai.