CMR:
\(A=2^1+2^2+...+2^{2010}⋮3,7\)
\(B=3^1+3^2+...+3^{2010}⋮4,13\)
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B = 3 + 3 2 + 3 3 + 3 4 + ... + 3 2010
B = ( 3 + 3 2 ) + ( 3 3 + 3 4 ) + ... + ( 3 2009 + 3 2010 )
B = ( 3 + 3 2 ) + ( 3 + 3 2 ) . 3 2 + ... + ( 3 + 3 2 ) . 3 2008
B = 12 + 12 . 3 2 + ... + 12 . 3 2008
B = 12 . ( 1 + 3 2 + ... + 3 2008 )
Vì 12 chia hết cho 4
=> 12 . ( 1 + 3 2 + ... + 3 2008 ) chia hết cho 4
=> B chia hết cho 4
B = 3 + 3 2 + 3 3 + 3 4 + ... + 3 2010
B = ( 3 + 3 2 + 3 3 ) + ( 3 4 + 3 5 + 3 6 ) + ... + ( 3 2008 + 3 2009 + 3 2010 )
B = ( 3 + 3 2 + 3 3 ) + ( 3 + 3 2 + 3 3 ) . 3 3 + ... + ( 3 + 3 2 + 3 3 ) . 3 2007
B = 39 + 39 . 3 3 + ... + 39 . 3 2007
B = 39 ( 1 + 3 3 + .... + 3 2007 )
Vì 39 chia hết cho 13
=> 39 ( 1 + 3 3 + .... + 3 2007 ) chia hết cho 13
=> B chia hết cho 13
trừi ơi , bạn có thôi ngay cái tính đó ko ,
bạn nói kiểu này , có khi bạn cần bài toán nào , bạn đăng lên ko ai làm đâu
Bài 2:
1: \(2A=2+2^2+...+2^{2011}\)
=>\(A=2^{2011}-1>B\)
2: \(A=\left(2010-1\right)\left(2010+1\right)=2010^2-1< B\)
3: \(A=1000^{10}\)
\(B=2^{100}=1024^{10}\)
mà 1000<1024
nên A<B
5: \(A=3^{450}=27^{150}\)
\(B=5^{300}=25^{150}\)
mà 27>25
nên A>B
A=2+2^2+2^3+2^4+...+2^2010
=(2+2^2)+(2^3+2^4)+...+(2^2009+2^2010)
=2(1+2)+2^3(1+2)+...+2^2009(1+2)
3(2+2^3+...+2^2009) chia hết cho 3
A=2+2^2+2^3+2^4+...+2^2010
=(2+2^2+2^3)+(2^4+2^5+2^6)+...+(2^2008+2^2009+2^2010)
=2(1+2+2^2)+2^4(1+2+2^2)+...+2^2008(1+2+2^2)
=7(2+2^4+...+2^2008) chia hết cho 7
A=(2+22)+(23+24)+...+(22009+22010)
A=2(1+2)+23(1+2)+...+22009(1+2)
A=2.3+23.3+...+22009.3
A=3(2+23+...+22009) chia hết cho 3
\(A=2^1+2^2+...+2^{2010}\)
\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(A=2.3+2^3.3+2^5.3+...+2^{2009}.3\)
\(A=3.\left(2+2^3+2^5+...+2^{2009}\right)\)\(⋮\)\(3\)
\(\Rightarrow A⋮3\)
\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2+4\right)+2^4\left(1+2+4\right)+2^7\left(1+2+4\right)+...+2^{2008}\left(1+2+4\right)\)
\(A=2.7+2^4.7+2^7.7+...+2^{2008}.7\)
\(A=7.\left(2+2^4+2^7+...+2^{2008}\right)\)\(⋮\)\(7\)
\(\Rightarrow A⋮7\)
\(B=3^1+3^2+...+3^{2010}\)
\(B=\left(3^1+3^2\right)+\left(3^3+3^4\right)+\left(3^5+3^6\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(B=3\left(1+3\right)+3^3\left(1+3\right)+3^5\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(B=3.4+3.3^3+3.3^5+...+3^{2009}.4\)
\(B=4.\left(3+3^3+3^5+...+3^{2009}\right)\)\(⋮\)\(4\)
\(\Rightarrow B⋮4\)
\(B=3^1+3^2+...+3^{2010}\)
\(B=\left(3^1+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\)
\(B=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^7\left(1+3+3^2\right)+...+3^{2008}\left(1+3+3^2\right)\)
\(B=3.13+3^4.13+3^7.13+...+3^{2008}.13\)
\(B=13.\left(3+3^4+3^7+...+3^{2008}\right)\)\(⋮\)\(13\)
\(\Rightarrow B⋮13\)