Cho `C = 1/3 – 2/3^2 + 3/3^3 – 4/3^4 … + 99/3^99 + 100/3^100` CMR : `C < 3/16` (Giúp nhanh nhé !) 18/07/2021 Bởi Caroline Cho `C = 1/3 – 2/3^2 + 3/3^3 – 4/3^4 … + 99/3^99 + 100/3^100` CMR : `C < 3/16` (Giúp nhanh nhé !)
Ta có : `C=1/3-2/3^2+3/3^3-4/3^4+…+99/3^99-100/3^100` `⇔ 3C=1-2/3+3/3^2-4/3^3+…+99/3^98-100/3^99` `⇔ 3C+C=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 `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-…-1/3^98` `⇔3A+A=3-1/3^99` `⇔A=(3-1/3^99)/4` `⇔ A=3/4-1/[4.3^99]` `⇔4C<3/4-1/[4.3^99]` `⇔C<(3/4-1/4.3^99)/4` `⇔C<3/16-1/[16.3^99]` `⇔C<3/16` Xin hay nhất ! Bình luận
Bạn tham khảo!
Ta có :
`C=1/3-2/3^2+3/3^3-4/3^4+…+99/3^99-100/3^100`
`⇔ 3C=1-2/3+3/3^2-4/3^3+…+99/3^98-100/3^99`
`⇔ 3C+C=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 `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-…-1/3^98`
`⇔3A+A=3-1/3^99`
`⇔A=(3-1/3^99)/4`
`⇔ A=3/4-1/[4.3^99]`
`⇔4C<3/4-1/[4.3^99]`
`⇔C<(3/4-1/4.3^99)/4`
`⇔C<3/16-1/[16.3^99]`
`⇔C<3/16`
Xin hay nhất !