Question

[cá mắm]

\(CMR:\) \(1-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{99}-\frac{1}{100}=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

Answer 1

[Huỳnh Quang Sang]

Biến đổi vp của đẳng thức :

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)

\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}-2\left[\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right]\)

\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{200}\)