Chứng minh `1/3-2/3^2+3/3^3-4/3^4+…+99/3^99-100/3^100<3/4`. Giúp em với! Mọi người cho em xin 1 slot thôi ạ! 11/07/2021 Bởi Vivian Chứng minh `1/3-2/3^2+3/3^3-4/3^4+…+99/3^99-100/3^100<3/4`. Giúp em với! Mọi người cho em xin 1 slot thôi ạ!
Đặt: `S = 1/3 – 2/3^2 + 3/3^3 – 4/3^4 + … + 99/3^99 – 100/3^100` `3S = 3 . ( 1/3 – 2/3^2 + 3/3^3 – 4/3^4 + … + 99/3^99 – 100/3^100)` `3S = 1 – 2/3 + 3/3^2 – 4/3^3 + … + 99/3^98 – 100/3^99` `3S + S = (1 – 2/3 + 3/3^2 – 4/3^3 + … + 99/3^98 – 100/3^99) + (1/3 – 2/3^2 + 3/3^3 – 4/3^4 + … + 99/3^99 – 100/3^100)` `4S = 1 – 2/3 + 3/3^2 – 4/3^3 + … + 99/3^98 – 100/3^99 + 1/3 – 2/3^2 + 3/3^3 – 4/3^4 + … + 99/3^99 – 100/3^100` `4S = 1 – 1/3 + 1/3^2 – 1/3^3 + 1/3^4 – … + 1/3^99 – 100/3^100` `=> 4S < 1 – 1/3 + 1/3^2 – 1/3^3 + 1/3^4 – … + 1/3^98 – 1/3^99` Lại có: Đặt: `A = 1 – 1/3 + 1/3^2 – 1/3^3 + 1/3^4 – … + 1/3^98 – 1/3^99` `3A = 3 . (1 – 1/3 + 1/3^2 – 1/3^3 + 1/3^4 – … + 1/3^98 – 1/3^99)` `3A = 3 – 1 + 1/3 – 1/3^2 + 1/3^3 – … + 1/3^97 – 1/3^98` `3A + A = (3 – 1 + 1/3 – 1/3^2 + 1/3^3 – … + 1/3^97 – 1/3^98) – (1 – 1/3 + 1/3^2 – 1/3^3 + 1/3^4 – … + 1/3^98 – 1/3^99)` `4A = 3 – 1/3^99` `=> 4S < 4A = 3 – 1/3^99` `4S < (3 – 1/3^99)/4` `4S < 3/4 – 1/(3^99 . 4)` `S < 3/4 – 1/(3^99 . 4)` `=> S < 3/4` Vậy `1/3 – 2/3^2 + 3/3^3 – 4/3^4 + … + 99/3^99 – 100/3^100 < 3/4` Bình luận
Đáp án+Giải thích các bước giải: `A=1/3-2/3^2+3/3^3-4/3^4+…+99/3^99-100/3^100` `⇔ 3A=1-2/3+3/3^2-4/3^3+…+99/3^98-100/3^99` `⇔ 3A+A=1-1/3+1/3^2-1/3^3+1/3^4-…+1/3^98-1/3^99-100/3^100 <1-1/3+1/3^2-1/3^3+1/3^4-…+1/3^98-1/3^99` Đặt `B=1-1/3+1/3^2-1/3^3+1/3^4-…+1/3^98-1/3^99` `⇔ 3B=3-1+1/3-1/3^2+1/3^3-…-1/3^98` `⇔3B+B=3-1/3^99` `⇔A=(3-1/3^99)/4` `⇔ A=3/4-1/[4.3^99]` `⇔4A<3/4-1/[4.3^99]` `⇔A<(3/4-1/4.3^99)/4` `⇔A<3/4-1/[4.3^99]` `⇔A<3/4` Bình luận
Đặt:
`S = 1/3 – 2/3^2 + 3/3^3 – 4/3^4 + … + 99/3^99 – 100/3^100`
`3S = 3 . ( 1/3 – 2/3^2 + 3/3^3 – 4/3^4 + … + 99/3^99 – 100/3^100)`
`3S = 1 – 2/3 + 3/3^2 – 4/3^3 + … + 99/3^98 – 100/3^99`
`3S + S = (1 – 2/3 + 3/3^2 – 4/3^3 + … + 99/3^98 – 100/3^99) + (1/3 – 2/3^2 + 3/3^3 – 4/3^4 + … + 99/3^99 – 100/3^100)`
`4S = 1 – 2/3 + 3/3^2 – 4/3^3 + … + 99/3^98 – 100/3^99 + 1/3 – 2/3^2 + 3/3^3 – 4/3^4 + … + 99/3^99 – 100/3^100`
`4S = 1 – 1/3 + 1/3^2 – 1/3^3 + 1/3^4 – … + 1/3^99 – 100/3^100`
`=> 4S < 1 – 1/3 + 1/3^2 – 1/3^3 + 1/3^4 – … + 1/3^98 – 1/3^99`
Lại có:
Đặt:
`A = 1 – 1/3 + 1/3^2 – 1/3^3 + 1/3^4 – … + 1/3^98 – 1/3^99`
`3A = 3 . (1 – 1/3 + 1/3^2 – 1/3^3 + 1/3^4 – … + 1/3^98 – 1/3^99)`
`3A = 3 – 1 + 1/3 – 1/3^2 + 1/3^3 – … + 1/3^97 – 1/3^98`
`3A + A = (3 – 1 + 1/3 – 1/3^2 + 1/3^3 – … + 1/3^97 – 1/3^98) – (1 – 1/3 + 1/3^2 – 1/3^3 + 1/3^4 – … + 1/3^98 – 1/3^99)`
`4A = 3 – 1/3^99`
`=> 4S < 4A = 3 – 1/3^99`
`4S < (3 – 1/3^99)/4`
`4S < 3/4 – 1/(3^99 . 4)`
`S < 3/4 – 1/(3^99 . 4)`
`=> S < 3/4`
Vậy `1/3 – 2/3^2 + 3/3^3 – 4/3^4 + … + 99/3^99 – 100/3^100 < 3/4`
Đáp án+Giải thích các bước giải:
`A=1/3-2/3^2+3/3^3-4/3^4+…+99/3^99-100/3^100`
`⇔ 3A=1-2/3+3/3^2-4/3^3+…+99/3^98-100/3^99`
`⇔ 3A+A=1-1/3+1/3^2-1/3^3+1/3^4-…+1/3^98-1/3^99-100/3^100 <1-1/3+1/3^2-1/3^3+1/3^4-…+1/3^98-1/3^99`
Đặt `B=1-1/3+1/3^2-1/3^3+1/3^4-…+1/3^98-1/3^99`
`⇔ 3B=3-1+1/3-1/3^2+1/3^3-…-1/3^98`
`⇔3B+B=3-1/3^99`
`⇔A=(3-1/3^99)/4`
`⇔ A=3/4-1/[4.3^99]`
`⇔4A<3/4-1/[4.3^99]`
`⇔A<(3/4-1/4.3^99)/4`
`⇔A<3/4-1/[4.3^99]`
`⇔A<3/4`