Đáp án:
$\\$
`B = 2^{2020}- 2^{2019} - 2^{2018} - ... - 2^2 - 2-1`
`-> B= 2^{2020} - (2^{2019} + 2^{2018} + ... + 2^2+2+1)` `(1)`
Đặt `A =1+2+... +2^{2018} + 2^{2019}`
`->2A = 2 (1+2+... +2^{2018} + 2^{2019})`
`-> 2A= 2 + 2^2 + ... + 2^{2019} + 2^{2020}`
`-> 2A - A = (2 + 2^2 + ... + 2^{2019} + 2^{2020}) - (1+2+... +2^{2018} + 2^{2019})`
`-> A = 2^{2020} - 1`
Với `A=2^{2020}-1` thay vào `(1)` ta được :
`-> B = 2^{2020} - (2^{2020}-1)`
`-> B = 2^{2020} - 2^{2020}+1`
`->B=(2^{2020}-2^{2020})+1`
`->B=1`
Vậy `B=1`