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!

Question

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!

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3 tuần 2021-07-11T14:07:48+00:00 2 Answers 2 views 0

1. 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

2. 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