Question

[Nam Lee]

Tính :

A = \(\dfrac{2}{60.63}+\dfrac{2}{63.66}+..........+\dfrac{2}{107.120}+\dfrac{2}{2006}\).

B = \(\dfrac{5}{40.44}+\dfrac{5}{44.48}+.....+\dfrac{5}{76.80}+\dfrac{5}{2006}\).

C = \(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\).

D = \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+.........+\dfrac{1}{2015.2018}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(=\dfrac{2}{3}\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}\right)+\dfrac{2}{2006}\)

\(=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)

\(=\dfrac{2}{3}\cdot\dfrac{1}{120}+\dfrac{1}{2003}=\dfrac{1}{180}+\dfrac{1}{2003}=\dfrac{2183}{180\cdot2003}\)

b: \(=\dfrac{5}{4}\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}\right)+\dfrac{5}{2006}\)

\(=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)

\(=\dfrac{5}{4}\cdot\dfrac{1}{80}+\dfrac{5}{2006}=\dfrac{1}{64}+\dfrac{5}{2006}=\dfrac{1163}{64192}\)

c: \(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+\dfrac{3}{17\cdot20}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{3}\cdot\dfrac{9}{20}=\dfrac{3}{20}\)