Question

[PHẠM NGỌC NGHỊ]

 bạn nào biết thì giúp mình với 

Answer 1

[Nguyễn Việt Lâm]

\(A=\dfrac{1}{2^1}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2023}{2^{2023}}\)

\(\Rightarrow2A=1+\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2023}{2^{2022}}\)

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 \(A< 2\)