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đầu bài là ê các bạn mình thấy lạ thật đấy chưa từng thấy đầu bài nào kì cục như vậy
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Ta có :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2009}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2010}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}+\frac{1}{2010}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}...+\frac{1}{2010}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1005}\right)\)
\(=\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2010}\)
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= ( 1 + 3 + 5 + ... + 2011 ) - ( 2 + 6 + 8 + ... + 2010 )
= [ 1006 x ( 2011 + 1 ) : 2 ] - [ 1005 x ( 2010 + 2 ) : 2 ]
= ( 1006 x 2012 : 2 ) - ( 1005 x 2012 : 2 )
= ( 2024072 : 2 ) - ( 2022060 : 2 )
= 1012036 - 1011030
= 1006
a = 1/2 nhân 2 + 1/3 nhân 3 + 1/4 nhân 4 + .....+ 1/2009 nhân 2009 + 1/2010 nhân 2010
so sánh a với 1
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a=1/2.2+1/3.3+1/4.4+...+1/2009.2009+1/2010.2010(có 2009 số hạng)
a=1+1+1+...+1+1(2009 số 1)
a=1.2009=2009
Vậy a>1
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Hình như đề bài phải là : Tính tổng : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}+\frac{1}{2010.2011}\)
Nếu thế giải như sau : \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}=\frac{2010}{2011}.\)Vậy tổng đó là 2010/2011.
Ta có :\(\frac{1}{1}:2+\frac{1}{2}:3+...+\frac{1}{2010}:2011\)
= \(\frac{1}{1}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+...+\frac{1}{2010}\times\frac{1}{2011}\)
= \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2010\times2011}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)
= \(1-\frac{1}{2011}\)
= \(\frac{2010}{2011}\)
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\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot....\cdot\frac{2009}{2010}\)
\(=\frac{\left(2\cdot3\cdot4\cdot5....\cdot2009\right)}{\left(2\cdot3\cdot4\cdot5\cdot....\cdot2009\right)\cdot2010}=\frac{1}{2010}\)
1/2x2/3x3/4x4/5x5/6x......x2009/2010
=(2x3x4x5x.....x2009)/(2x3x4x5x......x2009).2010=1/2010