Giúp mình nhé please
Câu 15
A= 1/2- 1/2^2+ 1/2^3- 1/2^4+ ...+ 1/2^2017- 1/2^2018
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Giải:
Ta có:A=1.2+2.3+3.4+...+2017.2018
3A=1.2.3 2.3.3+...+2017.2018.3
=1.2.(3-0)+2.3.(4-1)+...+2017.2018.(2019-2016)
=1.2.3+2.3.4+...+2017.2018.2019-1.2.0-2.3.1-...-2017.2018.1016
=2017.2018.2019-1.2.0
=2017.2018.2019
=>A=2017.2018.2019/3=2018.(2017.2019)/3
Và B=20183
/3=2018.2018.2018/3=2018.(2018.2018)/3
Lại có: 2017.2019=2017.(2018+1)=2017.2018+2017
2018.2018=(2017+1).2018=2017.2018+2018
Mà 2017.2018+2017<2017.2018+2018 =>2017.2019<2018.2018
=>2018.(2017.2019)<2018.(2018.2018)
=>A=2018.(2017.2019)/3<2018.(2018.2018)/3=B
=>A<B
Đặt \(S=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)
Biến đổi mẫu
\(\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}\)
\(=\left(2017+1\right)+\left(\frac{2016}{2}+1\right)+...+\left(\frac{1}{2017}+1\right)-2017\)
\(=2018+\frac{2018}{2}+...+\frac{2018}{2017}+\frac{2018}{2018}-2018\)
\(=2018.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)\)
\(\Rightarrow S=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}{2018.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}=\frac{1}{2018}\)