\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)

\(A=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{99\cdot100}\)

\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{4}-\frac{1}{100}\)

\(A=\frac{6}{25}\)

\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+.....+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}\\ =\frac{24}{100}=\frac{6}{25}\)

\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)

\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}-\frac{1}{6}+\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{4}-\frac{1}{100}\)

\(A=\frac{6}{25}\)

\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)

\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{4}-\frac{1}{100}\)

\(A=\frac{6}{25}\)

\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)

\(A=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{99.100}\)

\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)

Vậy A=6/25

\(A=\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+.....+\frac{1}{99\times100}\)

\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{4}-\frac{1}{100}\)

\(A=\frac{24}{100}=\frac{6}{25}\)

\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)

\(A=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{99\cdot100}\)

\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{4}-\frac{1}{100}=\frac{25}{100}-\frac{1}{100}=\frac{24}{100}=\frac{6}{25}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}\)

\(=\frac{24}{100}=\frac{6}{25}\)

\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)

\(\Rightarrow A=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{99.100}\)

\(\Rightarrow A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=\frac{1}{4}-\frac{1}{100}\)

\(\Rightarrow A=\frac{25}{100}-\frac{1}{100}\)

\(\Rightarrow A=\frac{24}{100}\)

\(\Rightarrow A=\frac{6}{25}\)

Bạn Ác Mộng làm đúng nhưng làm hơi tắt quá 

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{50}+\dfrac{1}{990}\)

\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{50}+\dfrac{1}{990}???\)

Quy luật của vế sau "..." sai, bạn xem lại đề bài!

Nếu đúng đề thì sẽ như sau:

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{9900}\)

Đề bài đúng là như vậy.

Giải:

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{9900}\)

\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{4}-\dfrac{1}{100}\)

\(=\dfrac{25-1}{100}\)

\(=\dfrac{24}{100}\)

\(=\dfrac{6}{25}\)

Sửa đề: \(\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{9702}+\frac{1}{9900}\)

Ta có: \(\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{9702}+\frac{1}{9900}\)

\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\cdots+\frac{1}{97\cdot98}+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)

\(=\frac14-\frac15+\frac15-\frac16+\cdots+\frac{1}{99}-\frac{1}{100}\)

\(=\frac14-\frac{1}{100}=\frac{24}{100}=\frac{6}{25}\)