Chứng minh rằng a/1/3+1/3^2+1/^3+…+1/3^99<1/2 b/1/3+1/3^2+1/3^3+...+1/3^100<3/4 24/08/2021 Bởi Emery Chứng minh rằng a/1/3+1/3^2+1/^3+…+1/3^99<1/2 b/1/3+1/3^2+1/3^3+...+1/3^100<3/4
` 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) ` Bình luận
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! Bình luận
` 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!