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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2+2^2+2^3+...+2^{20}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{19}+2^{20}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{19}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{19}\right)⋮3\)
\(A=2+2^2+2^3+...+2^{20}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{17}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{17}\right)⋮5\)
![](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)
vì 84 chia hết cho 3,nên 2+22+...+284 chia hết cho 3
vì 84 chia hết cho 7,nên 2+22+...+284 chia hết cho 7
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(A=3\left(2+2^3+...+2^{2009}\right)⋮3\)
\(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(A=7\left(2^1+2^4+...+2^{2008}\right)⋮7\)
Các ý dưới bạn làm tương tự nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
*Ta có: A\(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=\left(2+2^2\right)+2^2\times\left(2+2^2\right)+...+2^{2008}\times\left(2+2^2\right)\)
\(=\left(2+2^2\right)\times\left(1+2^2+2^3+...+2^{2008}\right)\)
\(=6\times\left(2^2+2^3+...+2^{2008}\right)\)
\(=3\times2\times\left(2^2+2^3+...+2^{2008}\right)\)
\(\Rightarrow A⋮3\)
*Ta có: A \(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=2\times\left(1+2+2^2\right)+2^4\times\left(1+2+2^2\right)+...+2^{2008}\times\left(1+2+2^2\right)\)
\(=\left(1+2+2^2\right)\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(=7\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(\Rightarrow A⋮7\)
Mình sửa lại đề C 1 chút xíu
*Ta có: C \(=3^1+3^2+3^3+3^4+...+3^{2010}\)
\(=\left(3+3^2\right)+3^2\times\left(3+3^2\right)+...+3^{2008}\times\left(3+3^2\right)\)
\(=\left(3+3^2\right)\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=12\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=4\times3\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(\Rightarrow C⋮4\)
Các câu khác làm tương tự nhé. Chúc bạn học tốt!
cho A 2 2 mũ 2 2 mũ 3 ...... 2 mũ 60chứng minh A chia hết cho 3,A chia hết cho 7 và A chi hết cho 42
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2+2^2+2^3+...+2^{60}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(A=2.3+2^3.3+...+2^{59}.3\)
\(A=3\left(2+2^3+...+2^{59}\right)\)
Vì \(3\left(2+2^3+...+2^{59}\right)⋮3\)
\(\Rightarrow A⋮3\)
\(A=2+2^2+2^3+...+2^{60}\)
\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(A=2.7+2^4.7+...+2^{58}.7\)
\(A=7\left(2+2^4+...+2^{58}\right)\)
Vì \(7\left(2+2^4+...+2^{58}\right)⋮7\)
\(\Rightarrow A⋮7\)
CM A chia hết 15
Ta có:
\(A=1+2+2^2+...+2^{23}\)
\(A=\left(1+2+2^2+2^3\right)+...+\left(2^{20}+2^{21}+2^{22}+2^{23}\right)\)
\(A=\left(1+2+2^2+2^3\right)+...+2^{20}\left(1+2+2^2+2^3\right)\)
\(A=15+...+2^{20}.15\)
\(A=15.\left(1+...+2^{20}\right)\) chia hết cho 15
=> đpcm
CM A chia hết cho 63:
Ta có:
\(A=1+2+2^2+...+2^{23}\)
\(A=\left(1+2+2^2+2^3+2^4+2^5\right)+...+\left(2^{18}+2^{19}+2^{20}+2^{21}+2^{22}+2^{23}\right)\)
\(A=63+...+2^{18}.63\)
\(A=63.\left(1+...+2^{18}\right)\) chia hết cho 63
=> đpcm