CMR:;1/3+2/3^2+3/3^3+….+100/3^100<3/4 Giúp mk vs. 23/08/2021 Bởi Kaylee CMR:;1/3+2/3^2+3/3^3+….+100/3^100<3/4 Giúp mk vs.
Đáp án: A = 1/3+2/3^2+3/3^3+….+100/3^100 (1) => 3A = 1 + 2/3 + 3/3^2 + …. + 100/3^99 (2) Lấy (2) – (1)ta đc 2A = 1 + 1/3 + 1/3^2 + … + 1/3^99 – 100/3^100 Đặt B = 1/3 + 1/3^2 + … + 1/3^99 (3) => 3B = 1 + 1/3 + …. + 1/3^98 (4) Lấy (4) – (3) ta đc 2B = 1 – 1/3^99 => B = 1 – 1/3^99 /2 = 1/2 – 1/3^99.2 => 2A = 1 + 1/2 – 3^99/2 + 100/3^99 < 1 + 1/2 = 3/2 => A < 3/2 : 2 = 3/4 Giải thích các bước giải: Bình luận
Đặt `A=1/3+2/3^2+3/3^3+…+100/3^100` `3A=1+2/3+3/3^2+…+100/3^99` `2A=1+2/3+3/3^2+…+100/3^99-(1/3+2/3^2+3/3^3+…+100/3^100)` `2A=1+2/3-1/3+3/3^2-2/3^2+…+100/3^99-99/3^99-100/3^100` `2A=1+1/3+1/3^2+1/3^3+…+1/3^99-100/3^100` `2A=1-100/3^100+(1/3+1/3^2+1/3^3+…+1/3^99)` Đặt `B=1/3+1/3^2+1/3^3+…+1/3^99` `3B=1+1/3+1/3^2+..+1/3^98` `2B=1+1/3+1/3^2+…1/3^98-(1/3+1/3^2+1/3^3+…+1/3^99)` `2B=1-1/3^99` `B=1/2-1/(3^99 .2)` `⇒2A=1-100/3^100+1/2-1/(3^99 .2)` `2A=1+1/2-100/3^100-1/(3^99 .2)` `⇒2A<1+1/2` `⇒A<3/4` Bình luận
Đáp án:
A = 1/3+2/3^2+3/3^3+….+100/3^100 (1)
=> 3A = 1 + 2/3 + 3/3^2 + …. + 100/3^99 (2)
Lấy (2) – (1)ta đc
2A = 1 + 1/3 + 1/3^2 + … + 1/3^99 – 100/3^100
Đặt B = 1/3 + 1/3^2 + … + 1/3^99 (3)
=> 3B = 1 + 1/3 + …. + 1/3^98 (4)
Lấy (4) – (3) ta đc
2B = 1 – 1/3^99
=> B = 1 – 1/3^99 /2 = 1/2 – 1/3^99.2
=> 2A = 1 + 1/2 – 3^99/2 + 100/3^99 < 1 + 1/2 = 3/2
=> A < 3/2 : 2 = 3/4
Giải thích các bước giải:
Đặt `A=1/3+2/3^2+3/3^3+…+100/3^100`
`3A=1+2/3+3/3^2+…+100/3^99`
`2A=1+2/3+3/3^2+…+100/3^99-(1/3+2/3^2+3/3^3+…+100/3^100)`
`2A=1+2/3-1/3+3/3^2-2/3^2+…+100/3^99-99/3^99-100/3^100`
`2A=1+1/3+1/3^2+1/3^3+…+1/3^99-100/3^100`
`2A=1-100/3^100+(1/3+1/3^2+1/3^3+…+1/3^99)`
Đặt `B=1/3+1/3^2+1/3^3+…+1/3^99`
`3B=1+1/3+1/3^2+..+1/3^98`
`2B=1+1/3+1/3^2+…1/3^98-(1/3+1/3^2+1/3^3+…+1/3^99)`
`2B=1-1/3^99`
`B=1/2-1/(3^99 .2)`
`⇒2A=1-100/3^100+1/2-1/(3^99 .2)`
`2A=1+1/2-100/3^100-1/(3^99 .2)`
`⇒2A<1+1/2`
`⇒A<3/4`