Question

[phuong phuong]

\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)=?

Answer 1

[Hồ Thu Giang]

\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

= \(\frac{1}{3}-\frac{1}{100}\)

= \(\frac{97}{300}\)