Đặt `B= 1 + 2 + 2^2 +...+ 2^2017 + 2^2018`
`2B = 2( 1 + 2 + 2^2 +...+ 2^2017 + 2^2018)`
`2B = 2 + 2^2 + 2^3 +...+ 2^2018 + 2^2019`
`2B -B =2 + 2^2 + 2^3 +...+ 2^2018 + 2^2019 - 1 -2 -2^2 -...-2^2017 - 2^2018`
`B = 2^2019 -1`
`=> A = 2^2019 - (2^2019-1)`
`=> A = 2^2019 - 2^2019 +1`
`=> A = 1`
Vậy `A= 1`