Câu hỏi của Bùi Xuân An

Question

Cho A=1+2+22+23+ ... +22018 , B=22019. Tính B - A

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Nguyễn Đăng Nhân · Accepted Answer

$A=1+2+2^2+...+2^{2018}$

$2A=2+2^3+2^4+...+2^{2019}$

$A=2A-A=1-2^{2019}$

$B-A=2^{2019}-\left(1-2^{2019}\right)$

$B-A=2^{2019}-1+2^{2019}$

$B-A=1$Câu trả lời: $A=1+2+2^2+...+2^{2018}$

$2A=2+2^3+2^4+...+2^{2019}$

$A=2A-A=1-2^{2019}$

$B-A=2^{2019}-\left(1-2^{2019}\right)$

$B-A=2^{2019}-1+2^{2019}$

$B-A=1$

『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』 · Answer

`#3107`$A=1+2+2^2+2^3+...+2^{2018}$ và $B=2^{2019}$Ta có:$A=1+2+2^2+2^3+...+2^{2018}$$2A=2+2^2+2^3+...+2^{2019}$$2A-A=\left(2+2^2+2^3+...+2^{2019}\right)-\left(1+2+2^2+2^3+...+2^{2018}\right)$$A=2+2^2+2^3+...+2^{2019}-1-2-2^2-2^3-...-2^{2018}$$A=2^{2019}-1$Vậy, $A=2^{2019}-1$Ta có:$B-A=2^{2019}-2^{2019}+1=1$Vậy, `B - A = 1.`Câu trả lời: `#3107`$A=1+2+2^2+2^3+...+2^{2018}$ và $B=2^{2019}$Ta có:$A=1+2+2^2+2^3+...+2^{2018}$$2A=2+2^2+2^3+...+2^{2019}$$2A-A=\left(2+2^2+2^3+...+2^{2019}\right)-\left(1+2+2^2+2^3+...+2^{2018}\right)$$A=2+2^2+2^3+...+2^{2019}-1-2-2^2-2^3-...-2^{2018}$$A=2^{2019}-1$Vậy, $A=2^{2019}-1$Ta có:$B-A=2^{2019}-2^{2019}+1=1$Vậy, `B - A = 1.`

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