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Ta có : \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)
\(\Rightarrow2A-A=3+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2017}}\)
\(=2-\frac{1}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)
\(A=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)
\(A=1+\left(1+\frac{2017}{2}\right)+\left(1+\frac{2016}{3}\right)+...+\left(1+\frac{1}{2018}\right)\)
\(A=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)
\(A=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)
Ta có: \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)
\(A=1+3^1+3^2+...+3^{2017}\)
\(3A=3+3^2+3^3+...+3^{2018}\)
\(3A-A=\left(3+3^2+3^3+...+3^{2018}\right)-\left(1+3^1+3^2+...+3^{2017}\right)\)
\(2A=3^{2018}-1\)
\(A=\frac{3^{2018}-1}{2}\)
\(\Rightarrow\)\(B-A=\frac{3^{2018}}{2}-\frac{3^{2018}-1}{2}=\frac{3^{2018}-3^{2018}+1}{2}=\frac{1}{2}\)
Vậy \(B-A=\frac{1}{2}\)
Chúc bạn học tốt ~
ta có: A = 1 + 31 + 32 + ...+ 32017
=> 3A = 31 + 32 + 33 + ....+ 32018
=> 3A - A = 32018 - 1
\(\Rightarrow A=\frac{3^{2018}-1}{2}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3^{2018-1}}{2}}{\frac{3^{2018}}{2}}=\frac{\frac{3^{2018}}{2}}{\frac{3^{2018}}{2}}-\frac{1}{\frac{3^{2018}}{2}}=1-\frac{1}{\frac{3^{2018}}{2}}\)
= (1 - 1/2).(1 - 1/3).(1 - 1/4) ... (1 - 1/2017)
=1/2 .2/3.3/4.......2016/1017
=1.2.3.4....2016/2.3.4.5...2017
=1.(2.3.4..2016)/(2.3.4..2016).2017
=1/2017( chia cả tử và mẫu cho 2.3.4.2016)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2016}{2017}.\frac{2017}{2018}\)
\(=\frac{1}{2018}\)
P/S: chúc bạn học tốt
bấm vào tìm kiếm sẽ có câu tương tự nha em