Question

\(\)Cho S = \(1\times2^0+2\times2^1+...+2019\times2^{2018}\) .  So sánh S với\(2^{2019}\times2018+2019\)

Answer 1

\(S=1.2^0+2.2^1+....+2018.2^{2017}+2019.2^{2018}\)

\(\Rightarrow2S=1.2^1+2.2^2+...+2018.2^{2018}+2019.2^{2019}\)

\(\Rightarrow S-2S=2^0+2^1+2^2+...+2^{2018}-2019.2^{2019}\)

\(\Rightarrow-S=2^0+2^1+...+2^{2018}-2019.2^{2019}\)

\(\Rightarrow-2S=2^1+2^2+...+2^{2019}-2019.2^{2020}\)

\(\Rightarrow-S-\left(-2S\right)=2^0-2020.2^{2019}+2019.2^{2020}\)

\(\Rightarrow S=1-1010.2^{2020}+2019.2^{2020}\)

\(\Rightarrow S=1019.2^{2020}-1=2038.2^{2019}-1\)

\(\Rightarrow S=2018.2^{2019}+20.2^{2019}-1>2018.2^{2019}+2019\)