Cho S = 2010+2010^2+2010^3 +....+2010^2009+2010^2010
Chứng tỏ S chia hết cho 2011
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dễ ợt
s=2010(1+20100+2010^3(1+2010)+............+2010^2009(1+2010)
s=2010.2011+2010^3.2011+.........+2010^2009.2011
s=2011(2010+2010^3+.......+2010^2009) chia hết cho 2011
\(S=\left(2010+2010^2\right)+\left(2010^3+2010^4\right)+...+\left(2010^{2009}+2010^{2010}\right)\)
\(S=2010\left(2010+1\right)+2010^3\left(2010+1\right)+...+2010^{2009}\left(2010+1\right)\)
\(S=2011.\left(2010+2010^3+2010^5+...+2010^{2009}\right)\) chia hết cho 2011
S = 2010 + 2010^2 + ........ + 2010^2010
= ( 2010 + 2010^2) + ....... + ( 2010^2009 + 2010^2010 )
= 2010. ( 1 + 2010 ) + .........+ 2010^2009. ( 1 + 2010 )
= 2010.2011 + ....... + 2010^2009.2011 chia hết cho 2011
=> S chia hết cho 2011
A = 20102011 - 20102010
A = 20102010 .( 2010 - 1)
A = 20102010.2009
2009 ⋮ 2009 ⇒ A = 20102010.2009 ⋮ 2009
Có : \(2009+2010>\dfrac{2009}{2010}\) ; \(2011+2012>\dfrac{2011}{2012}\)
\(\dfrac{2011}{2010}>1\) ; \(\dfrac{2010}{2011}< 1\) \(\Rightarrow\dfrac{2011}{2010}>\dfrac{2010}{2011}\)
Ta có : \(2009+2010+\dfrac{2011}{2010}+2011+2012>\dfrac{2009}{2010}+\dfrac{2010}{2011}+\dfrac{2011}{2012}\)
\(\Leftrightarrow B>A\)
Hay \(A< B\)
M= ( 1+20101)+(20102+20103)+(20104+20105)+(20106+20107)
M= 1.(2010+1) + 20122.(2010+1)+20104.(2010+1)+20106.(2010+1)
M= 2011.(1+20122+20104+20106)
Vậy M chia hết cho 2011
\(B=\dfrac{2008+2009+2010}{2009+2010+2011}=\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)Ta có : \(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)
\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)
\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)\(=>\dfrac{2008}{2009}+\dfrac{2009}{2010}+\dfrac{2010}{2011}>\dfrac{2008+2009+2010}{2009+2010+2011}\)
Hay A > B
S=(2010+2010^2)+(2010^3+2010^4)+...+(20010^2009)+(2010^2010)
=2010(1+2010)+2010^3(1+2010)+...+2010^2009(1+2010)
=2010.2011+2010^3.2011+...+2010^2009.2011
=2011(2010+...+2010^2009) chia hết 2011
nha