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

\(3A=3+3^2+3^3+...+3^{2018}\)

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

\(2A=3^{2018}-1\)

\(A=\frac{3^{2018}-1}{2}\)

\(\Rightarrow\)\(B-A=\frac{3^{2018}}{2}-\frac{3^{2018}-1}{2}=\frac{3^{2018}-3^{2018}+1}{2}=\frac{1}{2}\)

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

\(\Rightarrow A=\frac{3^{2018}-1}{2}\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{3^{2018-1}}{2}}{\frac{3^{2018}}{2}}=\frac{\frac{3^{2018}}{2}}{\frac{3^{2018}}{2}}-\frac{1}{\frac{3^{2018}}{2}}=1-\frac{1}{\frac{3^{2018}}{2}}\)

\(A=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)

\(A=1+\left(1+\frac{2017}{2}\right)+\left(1+\frac{2016}{3}\right)+...+\left(1+\frac{1}{2018}\right)\)

\(A=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)

\(A=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)

Ta có: \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)

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

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2016}{2017}.\frac{2017}{2018}\)

\(=\frac{1}{2018}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

A = 1 + 2 + 3 + ... + 2018

= ( 1 + 2018 ) + ( 2 + 2017) + ... + ( 1009 + 1010 )

= 2019 + 2019 + ... + 2019 ( có 1009 số 2019 )

= 2019 x 1009 = 2037171

B = 1 + 3 + 5 + ... + 2017

= ( 1 + 2017 ) + ( 3 + 2015 ) + ... + ( 1007 + 1010) + 1009 

= 2018 + 2018 + ... + 2018 + 1009 (có 504 số 2018)

= 2018 x 504 + 1009 = 1018081

Còn lại làm giống ý trên .