Toán 1.2-1/2! + 2.3-1/3! + 3.4/4! + … + 99.100-1/100! < 2 01/07/2021 By Alice 1.2-1/2! + 2.3-1/3! + 3.4/4! + … + 99.100-1/100! < 2
`(1.2 -1)/(2!) +(2.3 -1)/(3!)+(3.4)/(4!)+…+ (99.100-1)/(100!)` Đặt `A=(1.2 -1)/(2!) +(2.3 -1)/(3!)+(3.4)/(4!)+…+ (99.100-1)/(100!)` `=>A=(1.2)/(2!)- 1/(2!)+ (2.3)/(3!)-1/(3!)+…+(99.100)/(100!)-1/(100!)` `=>A=[((1.2)/(2!)+(2.3)/(3!)+….+(99.100)/(100!))-(1/(2!)+1/(3!)+…+1/(100!)]` `=>A=[(1+1+1/(2!)+….+1/(98!))-(1/(2!)+1/(3!)+…+1/(100!)]` `=>A=1-1/(99!)-1/(100!)` Mà `1-1/(99!)-1/(100!)<1=>1-1/(99!)-1/(100!)<2` `->đpcm` Trả lời
`A=1×2−1/2!+2×3−1/3!+3×4−1/4!+…+99×100−1/100!` `=1×2/2!−1/2!+2×3/3!−1/3!+…+99×100/100!−1/100!` `=(1×2/2!+2×3/3!+3×4/4!+…+99×100/100!)−(1/2!+1/3!+1/4!+…+1/100!)` `=(1+1+1/2!+…+1/98!)−(1/2!+1/3!+1/4!+…+1/100!)` `=1+1−1/99!−1/100!` `=2−1/99!−1/100!` $Mà:$ `2−199!−1100!<2` `⇒A<2` $#JAW#$ Trả lời
`(1.2 -1)/(2!) +(2.3 -1)/(3!)+(3.4)/(4!)+…+ (99.100-1)/(100!)`
Đặt `A=(1.2 -1)/(2!) +(2.3 -1)/(3!)+(3.4)/(4!)+…+ (99.100-1)/(100!)`
`=>A=(1.2)/(2!)- 1/(2!)+ (2.3)/(3!)-1/(3!)+…+(99.100)/(100!)-1/(100!)`
`=>A=[((1.2)/(2!)+(2.3)/(3!)+….+(99.100)/(100!))-(1/(2!)+1/(3!)+…+1/(100!)]`
`=>A=[(1+1+1/(2!)+….+1/(98!))-(1/(2!)+1/(3!)+…+1/(100!)]`
`=>A=1-1/(99!)-1/(100!)`
Mà `1-1/(99!)-1/(100!)<1=>1-1/(99!)-1/(100!)<2` `->đpcm`
`A=1×2−1/2!+2×3−1/3!+3×4−1/4!+…+99×100−1/100!`
`=1×2/2!−1/2!+2×3/3!−1/3!+…+99×100/100!−1/100!`
`=(1×2/2!+2×3/3!+3×4/4!+…+99×100/100!)−(1/2!+1/3!+1/4!+…+1/100!)`
`=(1+1+1/2!+…+1/98!)−(1/2!+1/3!+1/4!+…+1/100!)`
`=1+1−1/99!−1/100!`
`=2−1/99!−1/100!`
$Mà:$ `2−199!−1100!<2`
`⇒A<2`
$#JAW#$