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\(2+2^2+2^3+...+2^{11}+2^{12}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+\left(2^{10}+2^{11}+2^{12}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+2^7\left(1+2+2^2\right)+2^{10}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+2^7+2^{10}\right)\)chia hết cho \(7\).
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.
*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!
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)
a=2+2^2+2^3+...+2^10
a=(2+2^2)+(2^3+2^4)+...+(2^9+2^10)
a=2.(1+2)+2^3.(1+2)+...+2^9.(1+2)
a=3.(2+2^3+...+2^9)
=> a chia hết cho 3
a=2+2^2+2^3+...+2^10
a=(2+2^2+2^3+2^4+2^5)+(2^6+2^7+2^8+2^9+2^10)
a=2.(1+2+4+8+16)+2^6.(1+2+4+8+16)
a=31.(2+2^6)
=> a chia hết cho 31
chúc bạn học tốt nha
a,S=1+3+32+...+360
3S=3+32+33+...+361
3S-S=(3+32+33+...+361)-(1+3+32+...+360)
2S = 361 - 1
b,2S+1=361-1+1=361 = 3x-3
=>x-3=61=>x=64
c, S=1+3+32+...+360
=(1+3)+(32+33)+...+(359+360)
=4+32(1+3)+...+359(1+3)
=4+32.4+...+359.4
=4(1+32+...+359) chia hết cho 4
S=1+3+32+...+360
=(1+3+32)+....+(358+359+360)
=13+...+358(1+3+32)
=13+...+358.13
=13(1+...+358)
\(A=2+2^2+2^3+...+2^{10}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)
\(=3\left(2+2^3+...+2^9\right)⋮3\)
\(A=2+2^2+2^3+...+2^{10}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)\)
\(=\left(2+2^6\right).31⋮31\)
A=2+2^2+2^3+....+2^10:3
A=(2+2^2)+(2^3+2^4)+....+(2^9+2^10):3
A=2.(1+2)+2^3.(1+2)+...+2^9.(1+2):3
A=2.3+2^3.3+...+2^9.3:3
A=3.(2+2^3+...+2^9):3
vậy A:3