CHO A = 4 + 2^2 + 2^3 + … + 2^20. CHỨNG MINH RẰNG A CHIA HẾT CHO 128 04/09/2021 Bởi Autumn CHO A = 4 + 2^2 + 2^3 + … + 2^20. CHỨNG MINH RẰNG A CHIA HẾT CHO 128
Đáp án: $A = 4 + 2^2 + 2^3 + … + 2^20$ $A = 2^2 + 2^2 + 2^3 + … + 2^20$ $2A=2(2^2 + 2^2 + 2^3 + … + 2^20)$ $2A=2³+2³+2^4+2^5+…+2^21$ $2A-A=(2³+2³+2^4+2^5+…+2^21)-(2^2 + 2^2 + 2^3 + … + 2^20)$ $A=8=2^21-8$ $A=2^21$ Ta thấy $2^21$ chia hết cho $2^7$ Vậy $A$ chia hết cho $128$ Bình luận
Đặt A = 4 + $2^{2}$ + $2^{3}$ +…+ $2^{20}$ 2A = 8 + $2^{3}$ + $2^{4}$ +…+ $2^{21}$ 2A – A = (8 + $2^{3}$ + $2^{4}$ +…+ $2^{21}$) – (4 + $2^{2}$ + $2^{3}$ +…+ $2^{20}$) A = $2^{21}$ Mà 128 = $2^{7}$ => A ⋮ 128 Bình luận
Đáp án:
$A = 4 + 2^2 + 2^3 + … + 2^20$
$A = 2^2 + 2^2 + 2^3 + … + 2^20$
$2A=2(2^2 + 2^2 + 2^3 + … + 2^20)$
$2A=2³+2³+2^4+2^5+…+2^21$
$2A-A=(2³+2³+2^4+2^5+…+2^21)-(2^2 + 2^2 + 2^3 + … + 2^20)$
$A=8=2^21-8$
$A=2^21$
Ta thấy $2^21$ chia hết cho $2^7$
Vậy $A$ chia hết cho $128$
Đặt A = 4 + $2^{2}$ + $2^{3}$ +…+ $2^{20}$
2A = 8 + $2^{3}$ + $2^{4}$ +…+ $2^{21}$
2A – A = (8 + $2^{3}$ + $2^{4}$ +…+ $2^{21}$) – (4 + $2^{2}$ + $2^{3}$ +…+ $2^{20}$)
A = $2^{21}$
Mà 128 = $2^{7}$
=> A ⋮ 128