Question

[Thái Đào]

Tính giá trị biểu thức:

\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{4950}\)

Answer 1

[Nguyễn Thị Thảo]

\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{4950}\)

\(=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{9900}\)

\(=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{99.100}\)

\(=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)

\(=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)\(=2\left(\frac{1}{4}-\frac{1}{100}\right)\)

\(=2.\frac{6}{25}\)

\(=\frac{12}{25}\)