Tham khảo
`A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}`
Áp dụng công thức `\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}`
`⇒A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}`
`⇒A=1-\frac{1}{101}=\frac{100}{101}`
`\text{©CBT}`