Đáp án:
`B = 5^2/10^2 + 5^2/11^2 + ... + 5^2/99^2`
`->B = 5^2 × (1/10^2 + 1/11^2 + ... + 1/99^2)`
`-> B = 25 × (1/10^2 + 1/11^2 + ... + 1/99^2)`
`text{Vì}` \(\left\{ \begin{array}{l}\dfrac{1}{10^2} > \dfrac{1}{10 × 11}\\ \dfrac{1}{11^2} > \dfrac{1}{11 × 12}\\ ........\\ \dfrac{1}{99^2} > \dfrac{1}{99 × 100}\end{array} \right.\)
`text{Từ đó}`
`-> B > 25 × [1/(10 × 11) + 1/(11 × 12) + .... + 1/(99 × 100)]`
`-> B > 25 × [1/10 - 1/11 + 1/11 - 1/12 + .... + 1/99 - 1/100]`
`-> B > 25 × [1/10 + (- 1/11 + 1/11 - 1/12 + .... + 1/99) - 1/100]`
`-> B > 25 × [1/10 - 1/100]`
`-> B > 25 × 9/100`
`-> B > 9/4 (đpcm)`
