chứng tỏ rằng 1/2^2 + 1/3^2 + 1/4^2 + … + 1/2011^2 < 3/4 21/10/2021 Bởi Allison chứng tỏ rằng 1/2^2 + 1/3^2 + 1/4^2 + … + 1/2011^2 < 3/4
Giải thích các bước giải: `1/2^2 + 1/3^2 + 1/4^2 + … + 1/2011^2` `Ta` `thấy` `:` `1/2^2 = 1/4` ; `1/3² < 1/(2 . 3)` ; ………. ; `1/2011² < 1/2010.2011` `→ A < 1/(1.3) + 1/(2.3) + ….. + 1/(2010.2011)` `→ A < 1/2² + 1/2 – 1/3 + 1/3 – ….. + 1/2010 – 1/2011` `→ A < 1/4 + 1/2 – 1/2011` `→ A < 3/4 – 1/2011 < 3/4` `→ A < 3/4 (đpcm)` Bình luận
Đặt `A= 1/2^2 + 1/3^2 + 1/4^2 + … + 1/2011^2 ` `A<1/2^2+1/(2.3)+1/(3.4)+…+1/(2010.2011)` `A<1/2^2+1/2-1/3+1/3-1/4+…+1/2010-1/2011` `A<1/4+1/2-1/2011` `A<3/4-1/2011` `A<3/4 (đpcm)` Bình luận
Giải thích các bước giải:
`1/2^2 + 1/3^2 + 1/4^2 + … + 1/2011^2`
`Ta` `thấy` `:` `1/2^2 = 1/4` ; `1/3² < 1/(2 . 3)` ; ………. ; `1/2011² < 1/2010.2011`
`→ A < 1/(1.3) + 1/(2.3) + ….. + 1/(2010.2011)`
`→ A < 1/2² + 1/2 – 1/3 + 1/3 – ….. + 1/2010 – 1/2011`
`→ A < 1/4 + 1/2 – 1/2011`
`→ A < 3/4 – 1/2011 < 3/4`
`→ A < 3/4 (đpcm)`
Đặt `A= 1/2^2 + 1/3^2 + 1/4^2 + … + 1/2011^2 `
`A<1/2^2+1/(2.3)+1/(3.4)+…+1/(2010.2011)`
`A<1/2^2+1/2-1/3+1/3-1/4+…+1/2010-1/2011`
`A<1/4+1/2-1/2011`
`A<3/4-1/2011`
`A<3/4 (đpcm)`