A = 1 + 4 + 42 + 43 + ... + 41999 + 4 2000
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
\(1+3+3^2+..+3^{2000}=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+..+\left(3^{1998}+3^{1999}+3^{2000}\right)\)
\(=13.1+13\cdot3^3+..+13\cdot3^{1998}\) chia hết cho 13
tương tự
\(1+4+4^2+..+4^{2012}=\left(1+4+4^2\right)+..+\left(4^{2010}+4^{2011}+4^{2012}\right)\)
\(=21.1+21\cdot4^3+..+21.4^{2010}\) chia hết cho 21
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(42\cdot43+43\cdot57+43\right)-360:4\)
\(=43\cdot\left(42+57+1\right)-90\)
\(=42\cdot100-90\)
=4110
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+4+4^2+4^3+4^4+4^5+...+4^{2019}+4^{2020}+4^{2021}\)
\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{2019}+4^{2020}+4^{2021}\right)\)
\(=21+4^3\cdot21+...+4^{2019}\cdot21\)
\(=21\left(1+4^3+...+4^{2019}\right)⋮21\)
\(A=1+4+4^2+4^3+...+4^{2021}\\=(1+4+4^2)+(4^3+4^4+4^5)+(4^6+4^7+4^8)+...+(4^{2019}+4^{2020}+4^{2021})\\=21+4^3\cdot(1+4+4^2)+4^6\cdot(1+4+4^2)+...+4^{2019}\cdot(1+4+4^2)\\=21+4^3\cdot21+4^6\cdot21+...+4^{2019}\cdot21\\=21\cdot(1+4^3+4^6+...+4^{2019})\)
Vì \(21\cdot(1+4^3+4^6+...+4^{2019})\vdots21\)
nên \(A\vdots21\)
\(\text{#}Toru\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A) ( 42 x 43 + 43 x 57 + 43 ) - 360 : 4
= ( 42 x 43 + 43 x 57 + 43 x 1 ) - 360 : 4
= [ 43 x ( 42 + 57 + 1 ) ] - 360 : 4
= ( 43 x 100 ) - 360 : 4
= 4300 - 360 : 4
= 4300 - 90
= 4210
B) 322 - 19 x 4 + ( 981 : 9 - 13 )
= 322 - 19 x 4 + ( 109 - 13 )
= 322 - 19 x 4 + 96
= 322 - 76 + 96
= 246 + 96
= 342
C) 450 : 2 x 18 + 456 : 3 - 120
= 225 x 18 + 152 - 120
= 4050 + 152 - 120
= 4202 - 120
= 4082
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow4A=4+4^2+4^3+...+4^{100}\\ \Rightarrow4A-A=\left(4+4^2+4^3+...+4^{100}\right)-\left(1+4+4^2+...+4^{99}\right)\\ \Rightarrow3A=4^{100}-1< 4^{100}=B\\ \Rightarrow A< \dfrac{B}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(1+4+4^2\right)+...+4^{57}\left(1+4+4^2\right)\)
\(=21\left(1+...+4^{57}\right)⋮7\)
cứ tổng của 3 số liên tiếp được 1 số chia hết cho 7
=> (1+4+4^2)+(4^3+4^4+4^5)+.....+(4^57+4^58+4^59)(20 cặp số)
=> 21+ 4^3(1+4+4^2)+...+4^57(1+4+4^2)
......
Vì 21 chia hết cho 7=> 21.(........) chia hết cho 7=> A chia hết cho 7
đpcm
A = 1 + 4 + 42+ 43+ … + 41999+ 42000
= (1 + 4 + 42) + (43+ 44+ 45) + … + (41998+ 41999+ 42000)
= 21 + 43.(1 + 4 + 42) + … + 41998.(1 + 4 + 42)
= 21 + 43.21 + … + 41998.21
= 21.(1 + 43+ … + 41998)
đề bài yêu cầu gì vậy em?