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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!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2^{2019}-\left(2^{2018}+2^{2017}+2^{2016}+.....+2^1+2^0\right)\)
Đặt: \(B=2^{2018}+2^{2017}+2^{2016}+....+2^1+2^0\)
\(\Rightarrow2B=\left(2^{2018}+2^{2017}+2^{2016}+...+2^1+2^0\right)\)
\(\Rightarrow2B-B=\left(2^{2019}+2^{2018}+2^{2017}+...+2^2+2\right)-\left(2^{2018}+2^{2017}+2^{2016}+...+2^1+2^0\right)\)
\(\Rightarrow B=2^{2019}-1\)
\(\Rightarrow A=2^{2019}-\left(2^{2018}+2^{2017}+2^{2016}+.....+2^1+2^0\right)\)
\(=2^{2019}-\left(2^{2019}-1\right)=2^{2019}+2^{2019}+1>1\)
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Vì 2017<2018 nên\(\frac{1}{2017}\)>\(\frac{1}{2018}\)
⇒\(\frac{2}{2017}\)>\(\frac{1}{2018}\)
⇒\(\frac{2015}{2017}\)=1-\(\frac{2}{2017}\)<1-\(\frac{1}{2018}\)=\(\frac{2017}{2018}\)
Vậy, \(\frac{2015}{2017}\)< \(\frac{2017}{2018}\)
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