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Ta có :
\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\)
\(T=1+\frac{3}{1.2^2}+\frac{4}{2.2^2}+\frac{5}{2^2.2^2}+...+\frac{2016}{2^{2013}.2^2}+\frac{2017}{2^{1014}.2^2}\)
\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{4}+\frac{6}{8}+...+\frac{2016}{x}+\frac{2017}{x}\right)\)
\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2013}}+\frac{2017}{2^{2014}}\right)\)
Đến chỗ này chịu!
`Answer:`
\(T=\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\)
\(\Leftrightarrow2T=2+\frac{3}{2}+\frac{4}{2^2}+...+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\)
\(\Leftrightarrow2T-T=2+\left(\frac{3}{2}-\frac{2}{2}\right)+\left(\frac{4}{2^2}-\frac{4}{2^2}\right)+...+\left(\frac{2017}{2^{2015}}-\frac{2016}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)
\(\Leftrightarrow2T-T=2+\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)
Ta đặt \(V=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(\Rightarrow T=2+V-\frac{2017}{2^{2016}}\text{(*)}\)
\(\Leftrightarrow2V=1+\frac{1}{2}+...+\frac{1}{2^{2014}}\)
\(\Leftrightarrow2V-V=\left(1+\frac{1}{2}+...+\frac{1}{2^{2014}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)\)
\(\Leftrightarrow2V-V=1-\frac{1}{2^{2015}}\text{(**)}\)
Từ (*)(**)\(\Rightarrow T=2+\left(1-\frac{1}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)
\(\Leftrightarrow T=3-\frac{1}{2^{2015}}-\frac{2017}{2^{2016}}\)
`=>T<3`
