Question

[Lê Thị Xuân Niên]

1.Tính :

a ) \(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2018.2019}\)

b ) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{2017.2019}\)

c) \(S=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+....+\dfrac{1}{2018.2020}\)

d) \(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{2017.2018.2019}\)

2. Tính tổng:

a) \(S=1.2+2.3+3.4+...+2018.2019\)

b) \(S=3.5+5.7+7.9+...+2017.2019\)

c) \(S=2.4+4.6+6.8+...+2018.2020\)

d) \(S=1.2.3+2.3.4+3.4.5+...+2017.2018.2019\)

3.Tính

a ) \(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{2017.2020}\)

b ) \(S=\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+....+\dfrac{1}{2017.2019.2021}\)

Ai có công thức không cho mình xin với ????

Answer 1

[Phùng Khánh Linh]

Bài 1a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2018.2019}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2018}-\dfrac{1}{2019}\)

\(=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)

b) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)

\(2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2017.2019}\)

\(2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)

\(2S=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)

\(S=\dfrac{1009}{2019}\)

Còn lại bạn làm tương tự hết nhé .