D= $\frac{1}{2}$ ²+$\frac{1}{3}$ ²+$\frac{4}{2}$ ²+…+$\frac{10}{2}$ ² <1 08/09/2021 Bởi Emery D= $\frac{1}{2}$ ²+$\frac{1}{3}$ ²+$\frac{4}{2}$ ²+…+$\frac{10}{2}$ ² <1
Ta có: `1/2^2 = 1/2.2 < 1/1.2` `1/3^2 = 1/3.3 < 1/2.3` `1/4^2 = 1/4.4 < 1/3.4` Tương tự như vậy `=> 1/2^2 + 1/3^2 + 1/4^2+…+1/10^2 < 1/1.2 + 1/2.3+ 1/3.4+…+ 1/9.10` `=> D < 1 -1/2 +1/2-1/3+1/3-1/4+….+1/9-1/10` `=> D < 1-1/10 < 1` Vậy `D < 1` Bình luận
Ta có: `1/2^2 = 1/2.2 < 1/1.2`
`1/3^2 = 1/3.3 < 1/2.3`
`1/4^2 = 1/4.4 < 1/3.4`
Tương tự như vậy
`=> 1/2^2 + 1/3^2 + 1/4^2+…+1/10^2 < 1/1.2 + 1/2.3+ 1/3.4+…+ 1/9.10`
`=> D < 1 -1/2 +1/2-1/3+1/3-1/4+….+1/9-1/10`
`=> D < 1-1/10 < 1`
Vậy `D < 1`