` a) ` Đặt ` A = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^{99} `

Ta có:

` A = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^{99} `

` <=> 3A = 1 + 1/3 + 1/3^2 + ... + 1/3^{98} `

` <=> 3A - A = 1 + 1/3 + 1/3^2 + ... + 1/3^{98} - 1/3 - 1/3^2 - 1/3^3 - ... - 1/3^{99} `

` <=> 2A = 1 + (1/3 - 1/3) + (1/3^2 - 1/3^2) + ... + (1/3^{98} - 1/3^{98}) - 1/3^{99} `

` <=> 2A = 1 - 1/3^{99} `

` <=> A = (1 - 1/3^{99}) : 2 = 1/2 - \frac{1}{2.3^{99}} `

Do: ` 1/2 - \frac{1}{2.3^{99}} < 1/2 `

` => A < 1/2 ` ` (đpcm) `

` b) ` Đặt ` B = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^{100} `

Ta có:

` B = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^{100} `

` <=> 3B = 1 + 1/3 + 1/3^2 + ... + 1/3^{99} `

` <=> 3B - B = 1 + 1/3 + 1/3^2 + ... + 1/3^{99} - 1/3 - 1/3^2 - 1/3^3 - ... - 1/3^{100} `

` <=> 2B = 1 + (1/3 - 1/3) + (1/3^2 - 1/3^2) + ... + (1/3^{99} - 1/3^{99}) - 1/3^{100} `

` <=> 2B = 1 - 1/3^{100} `

` <=> B = (1 - 1/3^{100}) : 2 = 1/2 - \frac{1}{2.3^{100}} `

Do: ` 1/2 - \frac{1}{2.3^{100}} < 1/2 `

Mà ` 1/2 = 2/4 < 3/4 `

` => 1/2 - \frac{1}{2.3^{100}} < 3/4 `

` => B < 3/4 ` ` (đpcm) `

Chứng minh rằng:

a) 1/3 + 1/3^2 + 1/^3 + ... + 1/3^99 < 1/2

Ta đặt tổng 1/3 + 1/3^2 + 1/^3 + ... + 1/3^99 là A. Ta có:

A = 1/3 + 1/3^2 + 1/^3 + ... + 1/3^99

=> 3A = 1 + 1/3 + 1/3^2 + 1/^3 + ... + 1/3^98

=> 3A - A = 1 + 1/3 + 1/3^2 + 1/^3 + ... + 1/3^98 - 1/3 + 1/3^2 + 1/^3 + ... + 1/3^99

=> 2A = 1 + (1/3 - 1/3) + (1/3^2 - 1/3^2) + ... + (1/3^98 - 1/3^98) - 1/3^99

=> 2A = 1 - 1/ 3^99

=> A = (1 - 1/3^99) : 2 = 1/2 - 1/2.3^99

Vì 1/2 - 1/2.3^99 < 1/2

Vậy: A < 1/2 (Đpcm)

b) 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^100 < 3/4

* Vì mình không làm kịp nên Câu b bạn làm tương tự giống như cách làm Câu a nhé!