Question

[Hoàng Ngọc]

Tính A=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..........+\frac{1}{2017\cdot2018}\)

       B=\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+..........+\frac{1}{2015\cdot2017}\)

       C=\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+..........+\frac{1}{2017\cdot2020}\)

       Ai làm nhanh nhất mình sẽ Tick cho nha.

Answer 1

[Nguyễn Trung Nghĩa]

A = 

A = \(1-\frac{1}{2018}\)

A = \(\frac{2017}{2018}\)

Có : 

2.B = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)

2.B = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)

2.B = \(1-\frac{1}{2017}\)

2.B = \(\frac{2016}{2017}\)

B = \(\frac{2016}{2017}:2=\frac{1008}{2017}\)

Có :

3.C = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2017.2020}\)

3.C = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2017}-\frac{1}{2020}\)

3.C = \(\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)

C = \(\frac{2019}{2020}:3=\frac{673}{2020}\)

Answer 2

[Mori Ran]

a=1/1-1/2+1/2-1/3+...+1/2017-1/2018

=1/1-1/2018

=kq

may bai duoi lam tuong tu nha

mình chưa điền kết quả ban tu dien nha