Vì 2²⁰¹⁷ của Q < 2²⁰¹⁸ của P mà P còn được cộng nên P>Q

chả hiểu j cả

Ta có : 2²⁰¹⁸ > 2²⁰¹⁷ 

Nên : P = 1 + 2 + 2³ +.....+ 2²⁰¹⁸ > 2²⁰¹⁷

Vậy P > Q

\(A=2^{2019}-\left(2^{2018}+2^{2017}+2^{2016}+.....+2^1+2^0\right)\)

Đặt: \(B=2^{2018}+2^{2017}+2^{2016}+....+2^1+2^0\)

\(\Rightarrow2B=\left(2^{2018}+2^{2017}+2^{2016}+...+2^1+2^0\right)\)

\(\Rightarrow2B-B=\left(2^{2019}+2^{2018}+2^{2017}+...+2^2+2\right)-\left(2^{2018}+2^{2017}+2^{2016}+...+2^1+2^0\right)\)

\(\Rightarrow B=2^{2019}-1\)

\(\Rightarrow A=2^{2019}-\left(2^{2018}+2^{2017}+2^{2016}+.....+2^1+2^0\right)\)

\(=2^{2019}-\left(2^{2019}-1\right)=2^{2019}+2^{2019}+1>1\)

đoạn cuối cùng bạn làm sai rồi

Mình nhầm ạ ~

\(2^{2019}-\left(2^{2019}-1\right)=2^{2019}-2^{2019}+1=1.\)

Ta có:\(A=1+2+2^2+2^3+....+2^{2017}.\)

\(\Rightarrow2A=2^1+2^2+2^3+2^4+....+2^{2018}\)

\(\Rightarrow2A-A=\left(2^1+2^2+2^3+2^4+...+2^{2018}\right)-\left(1+2^1+2^2+2^3+...+2^{2017}\right)\)

\(\Rightarrow A=2^{2018}-1\)

\(\Rightarrow A=B\)

A =\(\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2018^2}\)

=> A <\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

=> A < \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)

=> A < \(\frac{1}{3}-\frac{1}{2019}\)

=> A < 1

=> A < 99 

3⁵¹>3⁵⁰=9²⁵>8²⁵=2⁷⁵>2⁷⁰

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, \(\frac{2015}{2017}\)< \(\frac{2017}{2018}\)