cho a = 2 mũ 0 + 2 mũ 1 + 2 mũ 2 + ... + 2 mũ 100 . chứng minh rằng a chia hết cho 3
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a=2+2^2+2^3+2^4+...+2^{100}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{99}\right)⋮3\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,A=2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)...+\left(2^{99}+2^{100}\right)\)
\(=6+2^2\cdot\left(2+2^2\right)+2^4\cdot\left(2+2^2\right)...+2^{98}\cdot\left(2+2^2\right)\)
\(=6+2^2\cdot6+2^4\cdot6...+2^{98}\cdot6\)
\(=6\cdot\left(1+2^2+2^4+...+2^{98}\right)\)
Vì \(6\cdot\left(1+2^2+2^4+...+2^{98}\right)⋮6\)
nên \(A⋮6\)
\(b,A=2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^3\right)+\left(2^2+2^4\right)+\left(2^3+2^5\right)+...+\left(2^{97}+2^{99}\right)+\left(2^{98}+2^{100}\right)\)
\(=10+2\cdot\left(2+2^3\right)+2^2\cdot\left(2+2^3\right)+...+2^{96}\cdot\left(2+2^3\right)+2^{97}\cdot\left(2+2^3\right)\)
\(=10+2\cdot10+2^2\cdot10+...+2^{96}\cdot10+2^{97}\cdot10\)
\(=10\cdot\left(1+2+2^2+...+2^{96}+2^{97}\right)\)
Vì \(10\cdot\left(1+2+2^2+...+2^{96}+2^{97}\right)⋮10\)
nên \(A⋮10\)
#\(Toru\)
![](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)
Ta có: A = 2 + 22 + 23 + 24 + ... + 299 + 2100
A = (2 + 22) + (23 + 24) + ... + (299 + 2100)
A = 6 + 22(2 + 22) + .... + 298(2 + 22)
A = 6 + 22.6 + ... + 298.6
A = 6.(1 + 22 + ... + 298) \(⋮\)6
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)
\(=7\left(2+...+2^{19}\right)⋮7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)
\(=7\left(2+...+2^{19}\right)⋮7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{19}\right)⋮7\)
A = 20 + 21 + 22 + ... + 2100
A = (20 + 21) + (22 + 23) + ...+ ( 299 + 2100)
A = (20 + 21) + 22 . (20 + 21) + ... + 299 . ( 20 + 21)
A = (20 + 21) . (20 + 22 + ... + 299)
A = 3 . (20 + 22 + ... + 299)
Vì 3 chia hết cho 3 nên 3 . (20 + 22 + ... + 299) chia hết cho 3.
=> A chia hết cho 3.