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S = 1/2 - 1/3 + 1/3 -1/4 + ......... +1/2011 -1/2012
S= 1/2 - 1/2012 = 1005/2012
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...-\frac{1}{2012}\)
\(S=\frac{1}{2}+0+0+0+...-\frac{1}{2012}\)
\(S=\frac{1}{2}-\frac{1}{2012}\)
\(S=\frac{1005}{2012}\)
\(A=\frac{2012}{1}\cdot\frac{1005}{2012}\)
\(A=1005\)
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\(\Leftrightarrow x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)=2011\)
\(\Leftrightarrow x\cdot\dfrac{2011}{2012}=2011\)
hay x=2012
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)x=2011\)
\(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)x=2011\)
\(\left(\dfrac{1}{1}-\dfrac{1}{2012}\right)x=2011\)
\(\dfrac{2011}{2012}x=2011\)
\(x=2012\)
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Xét vế trái biểu thức, ta có:
\(\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\right)\cdot x\)
\(=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\cdot x\)
\(=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\right]\cdot x\)
\(=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\right]\cdot x\)
\(=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot x\)
Xét vế phải biểu thức, ta có:
\(\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot2012\)
Từ đầu bài và 2 kết luận trên, ta suy ra:
\(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot x=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot2012\)
\(\Rightarrow x=2012\)
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\(A=\left(1-\frac{1}{2010}\right)-\left(1-\frac{1}{2011}\right)+\left(1-\frac{1}{2012}\right)-\left(1-\frac{1}{2013}\right)=-\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(A=-\frac{1}{2010.2011}-\frac{1}{2012.2013}\)
Vì 2010.2011 > 2009.2010 => \(\frac{1}{2010.2011}-\frac{1}{2009.2010}\)
\(-\frac{1}{2012.2013}>-\frac{1}{2011.2012}\)
=> A > B
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=> 3S = 1.2.3 + 2.3.3 + 3.4.3 + .... + 2011.2012.3
=> 3S = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + .... + 2011.2012.( 2013 - 2010 )
=> 3S = 1.2.3 + 2.3.4 - 1.2.3 + .... + 2011.2012.2013 - 2010.2011.2012
=> 3S = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 2010.2011.2012 - 2010.2011.2012 ) + 2011.2012.2013
=> 3S = 2011.2012.2013
=> S = ( 2011.2012.2013 ) : 3
3S=1.2.3+2.3.(4-1)+...............+2011.2012.(2013-2010)
3S=1.2.3+2.3.4-1.2.3+...............+2011.2012.2013-2010.2011.2012
3S=2011.2012.2013
S=2011.2012.2013:3
S=2714954572
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\(=2012.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2011.2012}\right)\)
\(=2012.\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{2012-2011}{2011.2012}\right)\)
\(=2012.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2011}-\frac{1}{2012}\right)\)
\(=2012.\left(1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{2011}+\frac{1}{2011}\right)-\frac{1}{2012}\right)\)
\(=2012.\left(1-\frac{1}{2012}\right)=\frac{2012.2011}{2012}=2011\)