Đáp án:
$\dfrac{333}{998}$
Giải thích các bước giải:
$\left(1-\dfrac{1}{1+2}\right)\!.\!\left(1-\dfrac{1}{1+2+3}\right)\!.\!\left(1-\dfrac{1}{1+2+3+4}\right)\!.\!.\!.\!\left(1-\dfrac{1}{1+2+3\ +\,.\!.\!.+\ 1996}\right)\\=\left(1-\dfrac{1}{3}\right)\!.\!\left(1-\dfrac{1}{6}\right)\!.\!\left(1-\dfrac{1}{10}\right)\!.\!.\!.\!\bigg(1-\dfrac{1}{\dfrac{(1996+1).1996}{2}}\bigg)\\=\dfrac23\cdot\dfrac56\cdot\dfrac9{10}\ \cdot\,.\!.\!.\cdot\left(1-\dfrac{2}{1996.1997}\right)\\=\dfrac46\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\ \cdot\,.\!.\!.\cdot\ \dfrac{1996.1997-2}{1996.1997}\\=\dfrac46\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\ \cdot\,.\!.\!.\cdot\ \dfrac{(1995+1).1997-2}{1996.1997}\\=\dfrac46\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\ \cdot\,.\!.\!.\cdot\ \dfrac{1995.1997+1997-2}{1996.1997}\\=\dfrac46\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\ \cdot\,.\!.\!.\cdot\ \dfrac{1995.1997+1995}{1996.1997}\\=\dfrac46\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\ \cdot\,.\!.\!.\cdot\ \dfrac{1995.(1997+1)}{1996.1997}\\=\dfrac46\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\ \cdot\,.\!.\!.\cdot\ \dfrac{1995.1998}{1996.1997}\\=\dfrac{1.4}{2.3}\cdot\dfrac{2.5}{3.4}\cdot\dfrac{3.6}{4.5}\ \cdot\,.\!.\!.\cdot\ \dfrac{1995.1998}{1996.1997}\\=\dfrac{1.4.2.5.3.6...1995.1998}{2.3.3.4.4.5...1996.1997}\\=\dfrac{(1.2.3...1995).(4.5.6...1998)}{(2.3.4...1996).(3.4.5...1997)}\\=\dfrac{1.1998}{1996.3}\\=\dfrac{333}{998}$
