Bài 21 :
Gọi `\frac{1}{3^2} + \frac{1}{4^2} + ... + \frac{1}{100^2}` là A
Ta có : `A < \frac{1}{2.3} + \frac{1}{3.4} + ... + \frac{1}{99.100} `
`A < \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{99} - \frac{1}{100}`
`A < \frac{1}{2} - \frac{1}{100} < \frac{1}{2}`
` => ĐPCM`