\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
= \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
= \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
= \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{99}-\frac{1}{100}\)