Question

3/4.3​+3/5.4​+3/5.6​+⋯+3/90.100

Answer 1

\(Sửa\dfrac{3}{4\cdot3}+\dfrac{3}{5\cdot4}+\dfrac{3}{5\cdot6}+...+\dfrac{3}{99\cdot100}\\ =3\cdot\left(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\right)\\ =3\cdot\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{3}-\dfrac{1}{100}\right)=1-\dfrac{3}{100}\\ =\dfrac{97}{100}\)