Question

bai 1:  s=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}\)

 

bai 2:  s=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\text{+}...\text{+}\frac{1}{2017.2019}\)

Answer 1

co ban nao ra chua de minh do ke qua coi dung ko?

Answer 2

\(S=\frac{1}{1.2}+\frac{1}{3.4}+.........+\frac{1}{199.200}\)

Answer 3

Bài 2:

\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2017.2019}\) 

\(=\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2017.2019}\right)\) 

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\right)\) 

\(=\frac{1}{2}\left(1-\frac{1}{2019}\right)\)

\(=\frac{1}{2}\cdot\frac{2018}{2019}=\frac{1009}{2019}\)