cho a= 2/5x7 + 2/8x10 + ... + 2/2015 x 2017 so sanh a voi 13%
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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)
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{2017.2017}\)
Ta có :
\(\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3.3}< \frac{1}{2.3}\)
\(\frac{1}{4.4}< \frac{1}{3.4}\)
........
\(\frac{1}{2017.2017}< \frac{1}{2016.2017}\)
=> \(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{2017.2017}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2016.2017}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}< 1\)
=> A < 1
\(a=\frac{1}{2.2}+\frac{1}{3.3}+........+\frac{1}{2017.2017}\)
\(a< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2016.2017}\)
\(a< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2016}-\frac{1}{2017}\)
\(a< 1-\frac{1}{2017}\)
Do \(a< 1-\frac{1}{2017}\)
\(\Rightarrow a< 1\)
![](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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![](https://rs.olm.vn/images/avt/0.png?1311)
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, \(\frac{2015}{2017}\)< \(\frac{2017}{2018}\)