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S = 1 + 3 + 32 + 33 + ... + 38 + 39
S = ( 1 + 3 ) + ( 32 + 33 ) + ... + ( 38 + 39 )
S = 4 + ( 1 . 32 + 3 .32 ) + .. + ( 1. 38 + 3 . 38 )
S = 4 + 4 .32 + .. + 4 . 38
S = 4 ( 1 + 32 + ... + 38 ) \(⋮\)4
Vậy S \(⋮\)4 ( đpcm )
Học tốt
#Dương
S = 1 + 3 + 32 + 33 + 34+35+ 36 + 37 + 38+39
S=( 1 + 3)+(32 + 33)+(34+35)+(36 + 37)+(38+39)
s=4+32.(3+1)+32.(3+1)+34.(3+1)+36.(3+1)+38.(3+1)
S=4.(1+32+34+36+38)
CHIA HẾT CHO 4
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(S=\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(=4\left(1+3^2+3^4+...+3^8\right)⋮4\)
b: \(S=\left(1+2\right)+2^2\left(1+2\right)+...+2^8\left(1+2\right)\)
\(=3\left(1+2^2+...+2^8\right)⋮3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3^1+3^2+3^3+...+3^8\)
\(=\left(3^1+3^2\right)+\left(3^3+3^4\right)+...+\left(3^7+3^8\right)\)
\(=1\left(3^1+3^2\right)+3^2\left(3^1+3^2\right)+...+3^6\left(3^1+3^2\right)\)
\(=1.12+3^2.12+...+3^6.12\)
\(=12.\left(1+3^2+...+3^6\right)⋮12\)
Ta có: \(7^6+7^5-7^4\)
\(=7^4\left(7^2+7-1\right)\)
\(=7^4.55\)
\(=7^4.5.11⋮11\)
BAN CO THe chi cach giai bai nay khong cach lam tri tiet minh 0 hieu
![](https://rs.olm.vn/images/avt/0.png?1311)
s=2+2^2+2^3+.....+2^100
s=2.(1+2+2^2+2^3)+......+2^97.(1+2+2^2+2^3)
s=2.15+....+2^97.15
s=15.(2+....+2^97)
=> s chia het cho 15
a=3+3^2+3^3+....+3^20
a=3.(1+3)+......+3^19.(1+3)
a=3.4+.....+3^19.4
a=4.(3+.....+3^19)
vay a chia het cho 4
![](https://rs.olm.vn/images/avt/0.png?1311)
32+33+34+...+3100
=9+9.3+9.32+...+9.398
=>tổng trên chia hết cho 9
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(C+3^{101}=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+.....+3^{96}\left(1+3+3^2\right)+3^{99}\left(1+3+3^2\right)\)
\(C+3^{101}=13+3^3.13+.....+3^{96}.13+3^{99}.13\)
=> C+3101 chia hết cho 13
Mặt khác 3101 không chia hết cho 13
=> C không chia hết cho 13
Ta có :
\(C=\left(1+7+7^2\right)+7^3\left(1+7+7^2\right)+....+7^{27}\left(1+3+3^2\right)+7^{30}\)
\(C=57+7^3.57+....+7^{27}.57+7^{30}\)
Mà 7^30 không chia hết cho 57
=> C không chia hết cho 57
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Đặt 3+3^2+3^3+ ... +3^99+3^100 là A
Ta có:
A = 3+3^2+3^3+ ... +3^99+3^100
A = (3+3^2)+(3^3+3^4)+ ... +(3^99+3^100)
A = 3.4 + 3^3.4 + ... + 3^99.4
A = 4.(3+3^3+...+3^99)
=> A chia hết cho 4
2) Để 35x7y chia hết cho 2; 5 => y = 0
Để 35x70 chia hết cho 3 => (3+5+x+7+0) chia hết cho 3 => (15+x) chia hết cho 3
=> x = 0;3;6;9
Vậy y = 0; x = 0; 3; 6; 9