Question

[Nguyễn Văn Tuấn]

giúp mik ik pls 

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)

\(=1-\dfrac{1}{6}=\dfrac{5}{6}\)

b: \(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{10100}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{100\cdot101}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}=\dfrac{100}{101}\)

c: \(A=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)

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

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

\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)

d: \(A=\dfrac{3}{10}+\dfrac{3}{40}+...+\dfrac{3}{340}\)

\(=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{17\cdot20}\)

\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\)

\(=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)