Câu 5:
\(B=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{43\cdot46}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{45}{46}=\dfrac{15}{23}\)