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\(A=1+3^1+3^2+...+3^{2017}\)
\(3A=3+3^2+3^3+...+3^{2018}\)
\(3A-A=\left(3+3^2+3^3+...+3^{2018}\right)-\left(1+3^1+3^2+...+3^{2017}\right)\)
\(2A=3^{2018}-1\)
\(A=\frac{3^{2018}-1}{2}\)
\(\Rightarrow\)\(B-A=\frac{3^{2018}}{2}-\frac{3^{2018}-1}{2}=\frac{3^{2018}-3^{2018}+1}{2}=\frac{1}{2}\)
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
\(\Rightarrow A=\frac{3^{2018}-1}{2}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3^{2018-1}}{2}}{\frac{3^{2018}}{2}}=\frac{\frac{3^{2018}}{2}}{\frac{3^{2018}}{2}}-\frac{1}{\frac{3^{2018}}{2}}=1-\frac{1}{\frac{3^{2018}}{2}}\)
\(A=1+3+3^2+3^3+3^4+....+3^{2016}\)
\(3A=3\left(1+3+3^2+3^3+3^4+....+3^{2016}\right)\)
\(3A=3+3^2+3^3+3^4+3^5+...+3^{2017}\)
\(3A-A=3^{2017}-1\)
\(2A=3^{2017}-1\)
\(A=\left(3^{2017}-1\right):2\)
\(M\text{à}\)\(A-B=\left(3^{2017}-1\right):2-3^{2017}:2\)
\(=3^{2017-1}=3^{2017}\)
\(=>A-B=3^{2017-1}-3^{2017}\)