Câu hỏi của PHẠM NGỌC NGHỊ

Question

bạn nào biết thì giúp mình với

Nguyễn Việt Lâm · Accepted Answer

$A=\dfrac{1}{2^1}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2023}{2^{2023}}$$\Rightarrow2A=1+\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2023}{2^{2022}}$Trừ vế cho vế:$\Rightarrow2A-A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}-\dfrac{2023}{2^{2023}}$$\Rightarrow A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}-\dfrac{2023}{2^{2023}}$$\Rightarrow2A=2+1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}-\dfrac{2023}{2^{2022}}$Trừ vế cho vế:$2A-A=2-\dfrac{2024}{2^{2022}}+\dfrac{2023}{2^{2023}}$$\Rightarrow A=2-\dfrac{1}{2^{2022}}\left(2024-\dfrac{2023}{2}\right)$$\Rightarrow A=2-\dfrac{2025}{2^{2023}}< 2$Vậy $A< 2$Câu trả lời: $A=\dfrac{1}{2^1}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2023}{2^{2023}}$$\Rightarrow2A=1+\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2023}{2^{2022}}$Trừ vế cho vế:$\Rightarrow2A-A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}-\dfrac{2023}{2^{2023}}$$\Rightarrow A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}-\dfrac{2023}{2^{2023}}$$\Rightarrow2A=2+1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}-\dfrac{2023}{2^{2022}}$Trừ vế cho vế:$2A-A=2-\dfrac{2024}{2^{2022}}+\dfrac{2023}{2^{2023}}$$\Rightarrow A=2-\dfrac{1}{2^{2022}}\left(2024-\dfrac{2023}{2}\right)$$\Rightarrow A=2-\dfrac{2025}{2^{2023}}< 2$Vậy $A< 2$

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP