Question

Answer 1

Câu 2:

\(=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)...\left(1+\dfrac{1}{98\cdot100}\right)\\ =\dfrac{1\cdot3+1}{1\cdot3}\cdot\dfrac{2\cdot4+1}{2\cdot4}\cdot\dfrac{3\cdot5+1}{3\cdot5}...\dfrac{98\cdot100+1}{98\cdot100}\\ =\dfrac{4}{1\cdot3}\cdot\dfrac{9}{2\cdot4}\cdot\dfrac{16}{3\cdot5}\cdot...\cdot\dfrac{9801}{98\cdot100}\\ =\dfrac{2\cdot3\cdot4\cdot...\cdot99}{1\cdot2\cdot3\cdot...\cdot98}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot99}{3\cdot4\cdot5\cdot...\cdot100}\\ =99\cdot\dfrac{2}{100}=\dfrac{198}{100}\)