Đáp án: $\dfrac{567}{760}$
Giải thích các bước giải:
Sửa lại đề bài: $\dfrac{3}{1.2.3}+\dfrac{3}{2.3.4}+...+\dfrac{3}{18.19.20}$
$\dfrac{3}{1.2.3}+\dfrac{3}{2.3.4}+...+\dfrac{3}{18.19.20}\\=\dfrac{3}{2}\left ( \dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{18.19.20} \right )\\=\dfrac{3}{2}\left ( \dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20} \right )\\=\dfrac{3}{2}\left ( \dfrac{1}{1.2}-\dfrac{1}{19.20} \right )\\=\dfrac{3}{2}\left ( \dfrac{1}{2}-\dfrac{1}{380} \right )\\=\dfrac{3}{2}.\dfrac{189}{380}=\dfrac{567}{760}$
