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NT
2
21 tháng 2 2021
Ta có: \(A=2+2^2+2^3+...+2^{120}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)
\(=14+2^3\cdot14+...+2^{117}\cdot14\)
\(=14\cdot\left(1+2^3+...+2^{117}\right)⋮7\)
Ta có: \(A=2+2^2+2^3+...+2^{120}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)
\(=62+2^5\cdot62+...+2^{115}\cdot62\)
\(=62\cdot\left(1+2^5+...+2^{115}\right)⋮31\)
Ta có: \(A=2+2^2+2^3+...+2^{120}\)
\(=\left(2+2^2+2^3+2^4+2^5+2^6\right)+\left(2^7+2^8+2^9+2^{10}+2^{11}+2^{12}\right)+...+\left(2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)
\(=126+126\cdot2^6+...+126\cdot2^{114}\)
\(=126\cdot\left(1+2^6+...+2^{114}\right)⋮21\)
VH
2
5 tháng 11 2021
\(A=\left(2+2^2+2^3\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\\ A=2\left(1+2^2+2^3\right)+...+2^{118}\left(1+2^2+2^3\right)\\ A=\left(1+2^2+2^3\right)\left(2+...+2^{118}\right)\\ A=7\left(2+...+2^{118}\right)⋮7\)
5 tháng 11 2021
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{118}\left(1+2+2^2\right)\)
\(=2.7+2^4.7+...+2^{118}.7=7\left(2+2^4+...+2^{118}\right)⋮7\)
19 tháng 12 2021
\(1,8^8+2^{20}=2^{24}+2^{20}=2^{20}\left(2^4+1\right)=2^{20}\cdot17⋮17\)
\(2,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\\ A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{119}\left(1+2\right)\\ A=3\left(2+2^3+...+2^{119}\right)⋮3\)
\(A=\left(2+2^2+2^3\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\\ A=2\left(1+2+2^2\right)+...+2^{118}\left(1+2+2^2\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{118}\right)=7\left(2+...+2^{118}\right)⋮7\\ A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{117}+2^{118}+2^{119}+2^{120}\right)\\ A=2\left(1+2+2^2+2^3\right)+...+2^{117}\left(1+2+2^2+2^3\right)\\ A=\left(1+2+2^2+2^3\right)\left(2+...+2^{117}\right)=15\left(2+...+2^{117}\right)⋮15\)
HP
1
DH
27 tháng 12 2023
Chứng minh chia hết cho 7
A = 21 + 22 + 23 + ................ + 2120
A = (21 + 22 + 23) + (24 + 25 + 26) + ................ + (2118 + 2119 + 2120)
A = 2.(1 + 2 + 4) + 24.(1 + 2 + 4) + ................. + 2118.(1 + 2 + 4)
A = 2.7 + 24 . 7 + ................ + 2118.7
A = 7.(2 + 24 + ........... + 2118)
AH
Akai Haruma
Giáo viên
31 tháng 12 2023
Câu 1:
$A=(2+2^2)+(2^3+2^4)+(2^5+2^6)+....+(2^{2019}+2^{2020})$
$=2(1+2)+2^3(1+2)+2^5(1+2)+....+2^{2019}(1+2)$
$=(1+2)(2+2^3+2^5+...+2^{2019})=3(2+2^3+2^5+...+2^{2019})\vdots 3$
-----------------
$A=2+(2^2+2^3+2^4)+(2^5+2^6+2^7)+....+(2^{2018}+2^{2019}+2^{2020})$
$=2+2^2(1+2+2^2)+2^5(1+2+2^2)+....+2^{2018}(1+2+2^2)$
$=2+(1+2+2^2)(2^2+2^5+....+2^{2018})$
$=2+7(2^2+2^5+...+2^{2018})$
$\Rightarrow A$ chia $7$ dư $2$.
AH
Akai Haruma
Giáo viên
31 tháng 12 2023
Câu 2:
$B=(3+3^2)+(3^3+3^4)+....+(3^{2021}+3^{2022})$
$=3(1+3)+3^3(1+3)+...+3^{2021}(1+3)$
$=(1+3)(3+3^3+...+3^{2021})=4(3+3^3+....+3^{2021})\vdots 4$
-------------------
$B=(3+3^2+3^3)+(3^4+3^5+3^6)+...+(3^{2020}+3^{2021}+3^{2022})$
$=3(1+3+3^2)+3^4(1+3+3^2)+....+3^{2020}(1+3+3^2)$
$=(1+3+3^2)(3+3^4+...+3^{2020})=13(3+3^4+...+3^{2020})\vdots 13$ (đpcm)
12 tháng 12 2021
Bài 1:
\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)
\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)
12 tháng 12 2021
Bài 2:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
