Toán chứng minh rằng : 1/6<1/5^2+1/6^2+1/7^2+...+1/100^2<1/4 13/09/2021 By Bella chứng minh rằng : 1/6<1/5^2+1/6^2+1/7^2+...+1/100^2<1/4
Tham khảo Đặt `A=\frac{1}{5^2}+\frac{1}{6^2}+…+\frac{1}{100^2}` Do đó:`\frac{1}{4.5}+\frac{1}{5.6}+…+\frac{1}{99.100}>A>\frac{1}{5.6}+\frac{1}{6.7}+…+\frac{1}{100.101}` `⇒\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+…+\frac{1}{99}-\frac{1}{100}>A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+…+\frac{1}{100}-\frac{1}{101}` `⇒\frac{1}{4}-\frac{1}{100}>A>\frac{96}{505}≈0,19>\frac{1}{6}≈0,17` Nên `\frac{1}{4}>A>\frac{1}{6}` Giải thích Áp dụng `\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}` `\text{©CBT}` Trả lời
Tham khảo
Đặt `A=\frac{1}{5^2}+\frac{1}{6^2}+…+\frac{1}{100^2}`
Do đó:`\frac{1}{4.5}+\frac{1}{5.6}+…+\frac{1}{99.100}>A>\frac{1}{5.6}+\frac{1}{6.7}+…+\frac{1}{100.101}`
`⇒\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+…+\frac{1}{99}-\frac{1}{100}>A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+…+\frac{1}{100}-\frac{1}{101}`
`⇒\frac{1}{4}-\frac{1}{100}>A>\frac{96}{505}≈0,19>\frac{1}{6}≈0,17`
Nên `\frac{1}{4}>A>\frac{1}{6}`
Giải thích
Áp dụng `\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}`
`\text{©CBT}`