\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\) => \(\frac{T}{2}=\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\)
=> \(T-\frac{T}{2}=\left(\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\right)-\left(\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\right)\)
<=> \(\frac{T}{2}=\frac{2}{2^1}+\left(\frac{3}{2^2}-\frac{2}{2^2}\right)+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+...+\left(\frac{2017}{2^{2016}}-\frac{2016}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)
<=> \(\frac{T}{2}=1+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)
Đặt: \(M=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}=>2M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\)
=> \(2M-M=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
=> \(M=\frac{1}{2}-\frac{1}{2^{2016}}< \frac{1}{2}\)
=> \(\frac{T}{2}< 1+\frac{1}{2}-\frac{2017}{2^{2017}}< 1+\frac{1}{2}=\frac{3}{2}\)
=> T < 3
