Toán so sánh tổng sau với 1 b= 1/2^2+1/3^2+1/4^2+…+1/100^2 21/11/2021 By Josephine so sánh tổng sau với 1 b= 1/2^2+1/3^2+1/4^2+…+1/100^2
`B= 1/(2^2) + 1/(3^2) + 1/(4^2)+…+ 1/(100^2)` $\text{Ta có}$ `1/(2^2) < 1/(1.2)` `1/(3^2) < 1/(2.3)` `1/(4^2) < 1/(3.4)` `…` `1/(100^2) < 1/(99.100)` `=> B < 1/(1.2) + 1/(2.3) + … + 1/(99.100)` `=> B < 1/1 – 1/2 + 1/2 – 1/3 +…+ 1/99 – 1/100` `=> B < 1/1 – 1/100` `=> B < 99/100` $\text{mà}$ `99/100 < 1` `=> B <1 (đpcm)` $\text{TEAM_IQ_2000}$ Trả lời
ta có:1/2^2=1/4;1/3^2<1/2.3=1/2-1/3;1/4^2<1/3.4=1/3-1/4 1/100^2<1/99.100=1/99-1/100 A=1/4+1/2-1/100<3/4 Trả lời
`B= 1/(2^2) + 1/(3^2) + 1/(4^2)+…+ 1/(100^2)`
$\text{Ta có}$
`1/(2^2) < 1/(1.2)`
`1/(3^2) < 1/(2.3)`
`1/(4^2) < 1/(3.4)`
`…`
`1/(100^2) < 1/(99.100)`
`=> B < 1/(1.2) + 1/(2.3) + … + 1/(99.100)`
`=> B < 1/1 – 1/2 + 1/2 – 1/3 +…+ 1/99 – 1/100`
`=> B < 1/1 – 1/100`
`=> B < 99/100`
$\text{mà}$ `99/100 < 1`
`=> B <1 (đpcm)`
$\text{TEAM_IQ_2000}$
ta có:1/2^2=1/4;1/3^2<1/2.3=1/2-1/3;1/4^2<1/3.4=1/3-1/4
1/100^2<1/99.100=1/99-1/100
A=1/4+1/2-1/100<3/4