Chứng minh rằng: $\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+…+\dfrac{1}{2017^3}<\dfrac{1}{2^2}$. Xin cảm ơn! 02/09/2021 Bởi Ariana Chứng minh rằng: $\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+…+\dfrac{1}{2017^3}<\dfrac{1}{2^2}$. Xin cảm ơn!
Đặt `A= 1/2^3 + 1/3^3 + 1/4^3+…+1/2017^3 ` Ta có: `1/2^3 = 1/(2.2.2) < 1/(1.2.3)` `1/3^3 = 1/(3.3.3) < 1/(2.3.4)` `…………………………………………………………….` `1/2017^3 = 1/(2017.2017.2017) < 1/(2016.2017.2018)` `=> 1/2^3 + 1/3^3 + 1/4^3+…+1/2017^3 < 1/(1.2.3) + 1/(2.3.4) +…+1/(2016.2017.2018)` `=> A < 1/2(1/1.2 – 1/2.3 + 1/2.3 – 1/3.4 +…+1/2016.2017 – 1/2017.2018` `=> A< 1/2( 1/1.2 – 1/2017.2018)` `=> A < 1/2 ( 1/2 – 1/2017.2018)` `=> A < 1/4 – 1/(2.2017.2018) < 1/4 = 1/2^2` Vậy `1/2^3 + 1/3^3 + 1/4^3+…+1/2017^3 < 1/2^2` Bình luận
Đặt `A= 1/2^3 + 1/3^3 + 1/4^3+…+1/2017^3 `
Ta có: `1/2^3 = 1/(2.2.2) < 1/(1.2.3)`
`1/3^3 = 1/(3.3.3) < 1/(2.3.4)`
`…………………………………………………………….`
`1/2017^3 = 1/(2017.2017.2017) < 1/(2016.2017.2018)`
`=> 1/2^3 + 1/3^3 + 1/4^3+…+1/2017^3 < 1/(1.2.3) + 1/(2.3.4) +…+1/(2016.2017.2018)`
`=> A < 1/2(1/1.2 – 1/2.3 + 1/2.3 – 1/3.4 +…+1/2016.2017 – 1/2017.2018`
`=> A< 1/2( 1/1.2 – 1/2017.2018)`
`=> A < 1/2 ( 1/2 – 1/2017.2018)`
`=> A < 1/4 – 1/(2.2017.2018) < 1/4 = 1/2^2`
Vậy `1/2^3 + 1/3^3 + 1/4^3+…+1/2017^3 < 1/2^2`