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Ta có \(B=\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{2}{2010}+1\right)+\left(\frac{1}{2011}+1\right)+1\)
\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2010}+\frac{2012}{2011}+\frac{2012}{2012}\)
\(B=2012.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)\)
B=2012.A
=>A/B=1/2012
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\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2011}\left(1+2+3+...+2011\right)\)
\(=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+...+\frac{1}{2011}\cdot\frac{2011.2012}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{2012}{2}\)
\(=\frac{2+3+4+...+2012}{2}\)
\(=\frac{\frac{2012\cdot2013}{2}-1}{2}=\frac{2025077}{2}\)
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Tổng các số tự nhiên từ 1 đến n là \(\frac{n\left(n+1\right)}{2}\)
Do đó \(A=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{2011}.\frac{2011.2012}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2012}{2}\)
\(=\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{2012}{2}\right)-\frac{1}{2}\)
\(=\frac{1+2+3+...+2012}{2}-\frac{1}{2}\)
\(=\frac{\frac{2012.2013}{2}}{2}-\frac{1}{2}\)
\(=1012538,5\)
Vậy ....
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Cậu ơi hình như đề bài đúng là:
P =\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2011}}{2011+\dfrac{2012}{2}+\dfrac{2009}{3}+...+\dfrac{1}{2011}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
S = \(1-\frac{1}{2^2}-\frac{1}{3^2}-....-\frac{1}{2011^2}<1-\frac{1}{2.3}-\frac{1}{3.4}-.....-\frac{1}{2011.2012}\)
Đặt A = \(-\frac{1}{2.3}-\frac{1}{3.4}-....-\frac{1}{2011.2012}=-\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2011.2012}\right)\)
\(A=-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{2012}\right)=-\left(\frac{1}{2}-\frac{1}{2012}\right)=-\frac{1005}{2012}\)
S = 1 + \(\left(-\frac{1005}{2012}\right)=\frac{1007}{2012}>\frac{1}{2011}\)
=> ĐPCM