Tính tổng \(2+2^2+2^3+...+2^{2011}\)
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Ta có A=\(2+2^2+2^3+...+2^{2011}\)
=>2A=\(2^2+2^3+2^4+...+2^{2012}\)
=>2A-A=\(\left(2^2+2^3+2^4+...+2^{2012}\right)-\left(2+2^2+2^3+...+2^{2011}\right)\)
=\(-2+2^2-2^2+2^3-2^3+2^4-2^4+...+2^{2011}-2^{2011}+2^{2012}\)
=\(2^{2012}-2\)
Vậy A=\(2^{2012}-2\)
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S=2+2^2+2^3+2^4+..........+2^2008
=> 2S= 2^2+2^3+2^4+..........+2^2008+2^2009
=> 2S-S = (2^2+2^3+2^4+..........+2^2008+2^2009)-(2+2^2+2^3+2^4+..........+2^2008)
=> S = 2^2009-2
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A=2+23+25+27+...+22011
=>4A=23+25+29+...+22013
=>3A=22013-2
=>A=(22013-2)/3
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a) -3 < x < 2 , x ∈ Z. Do đó : x ∈ { -2 ; -1 ; 0 ; 1 }
Tổng các số nguyên x là : -2 + (-1) + 0 + 1 = -2
b) -2011 < x < 2011; x ∈ Z. Do đó x ∈ { -2010 ; -2009 ; … ; 2009 ; 2010 }
Tổng các số nguyên x là : ( -2010 + 2010 ) + ( -2009 + 2009 ) + … + ( -1 + 1 ) + 0 = 0
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Ta có: 1-2+3-...+2011
= (1+3+...+2011)-(2+4+...+2010)
= [(2011+1).1006:2] - [(2010+2).1005:2]
= [2012.1006:2] - [2012.1005:2]
= 1006.1006 - 1006.1005
= 1006.(1006-1005)
= 1006.1 = 1006
Nhấn đúng cho mk nha!!!!!!!!!!!
=(1-2)+(3-4)+....+(2009-2010)+2011
= (-1) x 1005+ 2011
= 2011 - 1005
= 1006
\(A=2+2^2+2^3+...+2^{2011}\)
\(2A=2^2+2^3+2^4+...+2^{2012}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2012}\right)-\left(2+2^2+2^3+...+2^{2011}\right)\)
\(A=2^{2012}-2\)