\(T=\dfrac{1}{2^1}+\dfrac{2}{2^2}+...+\dfrac{2021}{2^{2021}}+\dfrac{2022}{2^{2022}}\)
\(\Leftrightarrow2T=1+\dfrac{1}{2}+\dfrac{3}{2^2}...+\dfrac{2020}{2^{2019}}+\dfrac{2021}{2^{2020}}+\dfrac{2022}{2^{2021}}\)
\(\Leftrightarrow2T-T=\left(1+\dfrac{1}{2}+\dfrac{3}{2^2}...+\dfrac{2020}{2^{2019}}+\dfrac{2021}{2^{2020}}+\dfrac{2022}{2^{2021}}\right)-\left(\dfrac{1}{2^1}+\dfrac{2}{2^2}+...+\dfrac{2021}{2^{2021}}+\dfrac{2022}{2^{2022}}\right)\)
\(\Leftrightarrow T=1+\dfrac{1}{2}+\dfrac{3}{2^2}...+\dfrac{2020}{2^{2019}}+\dfrac{2021}{2^{2020}}+\dfrac{2022}{2^{2021}}-\dfrac{1}{2^1}-\dfrac{2}{2^2}-...-\dfrac{2021}{2^{2021}}-\dfrac{2022}{2^{2022}}\)
\(\Leftrightarrow T=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2021}}-\dfrac{2022}{2^{2022}}\)
Đặt \(M=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2021}}\)
\(\Leftrightarrow2M=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\)
\(\Leftrightarrow2M-M=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2021}}\right)\)
\(\Leftrightarrow M=1-\dfrac{1}{2^{2021}}\)
Khi đó: \(T=1+M-\dfrac{2022}{2^{2022}}\)
\(\Leftrightarrow T=1+1-\dfrac{1}{2^{2021}}-\dfrac{2022}{2^{2022}}\)
\(\Leftrightarrow T=2-\left(\dfrac{1}{2^{2021}}+\dfrac{2022}{2^{2022}}\right)\)
\(Do\left(\dfrac{1}{2^{2021}}+\dfrac{2022}{2^{2022}}\right)>0\) \(nên\) \(suy\) \(ra\) \(T=2-\left(\dfrac{1}{2^{2021}}+\dfrac{2022}{2^{2022}}\right)< 2\)
Vậy \(T< 2\) (\(ĐPCM\))
