chứng minh : A = 2+2^2+2^3+2^4+.....+2^120 chia hết cho 7;31;17
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=2+2^2+...+2^{120}\)
\(\Rightarrow A=\left(2+2^2\right)+...+\left(2^{119}+2^{120}\right)\)
\(\Rightarrow A=\left(2+2^2\right)+...+2^{118}.\left(2+2^2\right)\)
\(\Rightarrow A=6+...+2^{118}.6\)
\(\Rightarrow A=6.\left(1+...+2^{118}\right)⋮3\Rightarrow A⋮3\left(đpcm\right)\)
b) \(A=2+2^2+...+2^{120}\)
\(\Rightarrow A=\left(2+2^2+2^3\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)
\(\Rightarrow A=\left(2+2^2+2^3\right)+...+2^{117}.\left(2+2^2+2^3\right)\)
\(\Rightarrow A=14+...+2^{117}.14\)
\(\Rightarrow A=14.\left(1+...+2^{117}\right)⋮7\Rightarrow A⋮7\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 21 + 22 + 23 + ................ + 2120
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)
Chứng minh chia hết cho 31
A = 21 + 22 + 23 + ................ + 2120
A = (21 + 22 + 23 + 24 + 25) + (26 + 27 + 28 + 29 + 210) + ................ + (2116 + 2117 + 2118 + 2119 + 2120)
A = 2.(1 + 2 + 4 + 8 + 16) + 26.(1 + 2 +4 + 8 + 16) + ............. + 2116.(1 + 2 + 4 + 8 + 16)
A = 2.31 + 26.31 + ....... + 2116 . 31
A = 31.(2 + 26 + ........... + 2116)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 2 + 22 + ... + 2120
Chứng minh chia hết cho 3
A = ( 2 + 22 ) + ( 23 + 24 ) + ... + ( 2119 + 2120 )
= 2( 1 + 2 ) + 23( 1 + 2 ) + ... + 2119( 1 + 2 )
= 2.3 + 23.3 + ... + 2119.3
= 3( 2 + 23 + ... + 2119 ) chia hết cho 3 ( đpcm )
Chứng minh chia hết cho 7
A = ( 2 + 22 + 23 ) + ( 24 + 25 + 26 ) + ... + ( 2118 + 2119 + 2120 )
= 2( 1 + 2 + 22 ) + 24( 1 + 2 + 22 ) + ... + 2118( 1 + 2 + 22 )
= 2.7 + 24.7 + ... + 2118.7
= 7( 2 + 24 + ... + 2118 ) chia hết cho 7 ( đpcm )
Chứng minh chia hết cho 15
A = ( 2 + 22 + 23 + 24 ) + ( 25 + 26 + 27 + 28 ) + ... + ( 2117 + 2118 + 2119 + 2120 )
= 2( 1 + 2 + 22 + 23 ) + 25( 1 + 2 + 22 + 23 ) + ... + 2117( 1 + 2 + 22 + 23 )
= 2.15 + 25.15 + ... + 2117.15
= 15( 2 + 25 + ... + 2117 ) chia hết cho 15 ( đpcm )
1) Ta có: \(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{119}\left(1+2\right)\)
\(A=3\left(2+2^3+...+2^{119}\right)\) chia hết cho 3
2) Ta có: \(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{118}\left(1+2+2^2\right)\)
\(A=7\left(2+2^4+...+2^{118}\right)\) chia hết cho 7
3) Ta có: \(A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{117}+2^{118}+2^{119}+2^{120}\right)\)
\(A=2\left(1+2+2^2+2^3\right)+...+2^{117}\left(1+2+2^2+2^3\right)\)
\(A=15\left(2+2^5+...+2^{117}\right)\) chia hết cho 15
![](https://rs.olm.vn/images/avt/0.png?1311)
A=2+2^2+2^3+...+2^120
A=(2+2^2+2^3)+(2^4+2^5+2^6)...+(2^118+2^119+2^120)
A=2.(1+2+2^2)+2^4(1+2+2^2)+2^118(1+2+2^2)
A=2.7+2^4.7+...+2^118.7
Ta có A=2.7+2^4.7+...+2^118.7 chia hết cho 7
=>A=2+2^2+2^3+...+2^120 chia hết cho 7
A=2+2^2+...+2^120
=(2+2^2+2^3+2^4+2^5+2^6)+(2^7+2^8+2^9+2^10+2^11+2^12)+.....+(2^120+2^119+2^118+2^117+2^116+2^115)
=2(1+2+2^2+2^3+2^4+2^5)+2^7(1+2+2^2+2^3+2^4+2^5)+.....+2^115(1+2+2^2+2^3+2^4+2^5)
=2*63+2^7*63+...+2^115*63
=63(2+2^7+...+2^115) Vì 63 chia hết cho 7=>63(2+2^7+..+2^115) chia hết cho 7
=>A chia hết cho 7
![](https://rs.olm.vn/images/avt/0.png?1311)
c)D=4+42+43+44+...+42012
D=(4+42)+(43+44)+...+(42011+42012)
D=4.5+43.5+45.5+...+42011.5
D=5.(4+43+42011)
=>D chia hết cho 5
=>ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)
\(=57\left(7+7^4+...+7^{118}\right)⋮57\)
\(A=7\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)
\(=57\left(7+...+7^{118}\right)⋮57\)