A=1/2^2+1/3^2+1/4^2+…+1/9^2 Chứng tỏ 8/9>A>2/5 21/09/2021 Bởi Charlie A=1/2^2+1/3^2+1/4^2+…+1/9^2 Chứng tỏ 8/9>A>2/5
`\text{Ta có:}` `A = 1/2^2 + 1/3^2 + 1/4^2 + … + 1/9^2 < 1/2^2 + 1/2 . 3 + 1/3 . 4 + … + 1/8 . 9` `\text{Mà:}` `1/2^2 + 1/2 . 3 + 1/3 . 4 + … + 1/8 . 9 = 1/2^2 + 1/2 -1/3 + 1/3 – 1/4 + … + 1/8 – 1/9` `= 1/4 + 1/2 – 1/9 = 23/36 < 8/9` `\text{Ta có:}` `A = 1/2^2 + 1/3^2 + 1/4^2 + … + 1/9^2 > 1/2^2 + 1/3 . 4 + 1/4 . 5 + … + 1/9 . 10` `\text{Mà:}` `1/2^2 + 1/3 . 4 + 1/4 . 5 + … + 1/9 . 10 = 1/2^2 + 1/3 – 1/4 + 1/4 – 1/5 + … + 1/9 – 1/10` `= 1/2^2 + 1/3 – 1/10 = 19/20 > 2/5` `⇒ 8/9 > A > 2/5` Bình luận
Tham khảo Xét `A` `⇒\frac{1}{1.2}+\frac{1}{2.3}+…+\frac{1}{8.9}>A>\frac{1}{2.3}+\frac{1}{3.4}+…+\frac{1}{9.10}` Áp dụng `\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}` `⇒1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+…+\frac{1}{8}-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{9}-\frac{1}{10}` `⇒1-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{10}` `⇒\frac{8}{9}>A>\frac{2}{5}` `\text{©CBT}` Bình luận
`\text{Ta có:}` `A = 1/2^2 + 1/3^2 + 1/4^2 + … + 1/9^2 < 1/2^2 + 1/2 . 3 + 1/3 . 4 + … + 1/8 . 9`
`\text{Mà:}` `1/2^2 + 1/2 . 3 + 1/3 . 4 + … + 1/8 . 9 = 1/2^2 + 1/2 -1/3 + 1/3 – 1/4 + … + 1/8 – 1/9`
`= 1/4 + 1/2 – 1/9 = 23/36 < 8/9`
`\text{Ta có:}` `A = 1/2^2 + 1/3^2 + 1/4^2 + … + 1/9^2 > 1/2^2 + 1/3 . 4 + 1/4 . 5 + … + 1/9 . 10`
`\text{Mà:}` `1/2^2 + 1/3 . 4 + 1/4 . 5 + … + 1/9 . 10 = 1/2^2 + 1/3 – 1/4 + 1/4 – 1/5 + … + 1/9 – 1/10`
`= 1/2^2 + 1/3 – 1/10 = 19/20 > 2/5`
`⇒ 8/9 > A > 2/5`
Tham khảo
Xét `A`
`⇒\frac{1}{1.2}+\frac{1}{2.3}+…+\frac{1}{8.9}>A>\frac{1}{2.3}+\frac{1}{3.4}+…+\frac{1}{9.10}`
Áp dụng `\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}`
`⇒1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+…+\frac{1}{8}-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{9}-\frac{1}{10}`
`⇒1-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{10}`
`⇒\frac{8}{9}>A>\frac{2}{5}`
`\text{©CBT}`