Question

tiếp help A=\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+.....+\(\dfrac{1}{99.101}\) help me :)

Answer 1

\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\left(\dfrac{100}{101}\right)=\dfrac{50}{101}\)

Answer 2

\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)

Answer 3

= 1/2 . (1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ...+ 1/99 + 1/101)

= 1/2 . (1/1 - 1/101)

= 1/2 . 100/101

= 50/101

 

Answer 4

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{99.101}\)

\(2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\)

\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(2A=\dfrac{101}{101}-\dfrac{1}{101}\)

\(2A=\dfrac{100}{101}\)

\(A=\dfrac{50}{101}\)