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\(B=2+2^2+2^3+...+2^{30}\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+...+2^{29}\left(1+2\right)\)
\(=2\cdot3+2^3\cdot3+2^5\cdot3+...+2^{29}\cdot3\)
\(=3\left(2+2^3+2^5+...+2^{29}\right)⋮3\)
Mặt khác:\(B=2+2^2+2^3+...+2^{30}\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+....+2^{28}\left(1+2+2^2\right)\)
\(=2\cdot7+2^4\cdot7+....+2^{28}\cdot7\)
\(=7\left(2+2^4+...+2^{28}\right)⋮7\)
Mà (3;7)=1
\(\Rightarrow B⋮3\cdot7=21\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=2+2^2+...........+2^60
c\m c\h cho 3:2+2^2+....+2^60=2.(1+2)+........+2^59(1+2)
=2.3+.........+2^59.3
=(2+...+2^59).3
=>A chia hết cho 3
cau tiếp tuong tu
3
Ta chứng minh A chia hết cho 3:
A=(2+2^2)+(2^3+2^4)+...+(2^59+2^60)
=2.(1+2)+2^3.(1+2)+...+2^59.(1+2)
=2.3+2^3.3+...+2^59.3
=3.(2+2^3+...+2^59) chia hết cho 3
Ta chứng minh A chia hết cho 7
A=(2+2^2+2^3)+(2^4+2^5+2^6)+...+(2^58+2^59+2^60)
=2.(1+2+4)+2^4.(1+2+4)+...+2^58.(1+2+4)
=2.7+2^4.7+...+2^58.7
=7.(2+2^4+...+2^58) chia hết cho 7
Ta chứng minh A chia hết cho 15
A=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+...+(2^57+2^58+2^59+2^60)
=2.(1+2+4+8)+2^5.(1+2+4+8)+....+2^57.(1+2+4+8)
=2.15+2^5.15+..+2^57.15
=15.(2+2^5+...+2^57) chia hết cho 15
![](https://rs.olm.vn/images/avt/0.png?1311)
112 chia hết cho 7
nhưng : 2.1+3.1=5 không chia hết cho 7
![](https://rs.olm.vn/images/avt/0.png?1311)
B = 31 + 32 + 33 + ... + 328 + 329 + 330
B = ( 31 + 32 + 33 ) + ... + ( 328 + 329 + 330 )
B = 31 . ( 1 + 3 + 32 ) + ... + 328 . ( 1 + 3 + 32 )
B = 31 . 13 + ... + 328 . 13
B = 13 . ( 3 + ... + 328 ) \(⋮\)13
Vậy B \(⋮\)13 ( dpcm )
\(B=3^1+3^2+3^3+3^4+3^5+............+3^{30}\)
\(\Rightarrow B=\left(3^1+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+............+\left(3^{28}+3^{29}+3^{30}\right)\)
\(\Rightarrow B=3^1.\left(1+3+3^2\right)+3^4.\left(1+3+3^2\right)+.........+3^{28}.\left(1+3+3^2\right)\)
\(\Rightarrow B=3^1.13+3^4.13+.........+3^{28}.13\)
\(\Rightarrow B=13\left(3^1+3^4+.........+3^{28}\right)\)
Mà 13 \(⋮\)13 \(\Rightarrow13\left(3^1+3^4+...........+3^{28}\right)⋮13\)
Vậy B chia hết cho 13
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=1+3+3^2+3^3+...+3^{98}+3^{99}\)
\(=\left(1+3+3^2+3^3\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)
\(=40+...+3^{96}.\left(1+3+3^2+3^3\right)\)
\(=40+...+3^{96}.40\)
\(=\left(1+...+3^{96}\right).40⋮40\)
\(\Rightarrow\) \(B⋮40\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì 13 là lẻ \(\Rightarrow\) 13, 132, 133, 134, 135, 136 là lẻ.
Mà lẻ + lẻ + lẻ + lẻ + lẻ + lẻ = chẵn nên 13 + 132 + 133 + 134 + 135 + 136 là chẵn. \(\Rightarrow\) 13 + 132 + 133 + 134 + 135 + 136 \(⋮\) 2
\(\Rightarrow\) ĐPCM
Ta có:
B=1+2+22+23+...+21499
B=(1+2)+(22+23)+...+(21498+21499)
B=3+22(1+2)+...+21498(1+2)
B=3+22.3+...+21498.3
B=3(1+22+...+21498)
\(\Rightarrow B⋮3\)
Vậy\(B⋮3\)