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a, \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2012\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
b, \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}=\frac{1}{2017}\)
Câu 2: Ta có \(S=6^2+18^2+30^2+...+126^2\)
\(S=6^2\left(1^2+3^2+5^2+...+21^2\right)\)
\(=6^2.1771=36.1771=63756\)
\(S=1+\frac{1}{1+2}+\frac{1}{1+2+3}+..+\frac{1}{1+2+3+..+2011}\)
\(S=1+\frac{1}{2.\left(2+1\right):2}+\frac{1}{3.\left(3+1\right):2}+...+\frac{1}{2011.\left(2011+1\right):2}\)
\(S=1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2011.2012}\)
\(S=1+2\left(\frac{1}{2}-\frac{1}{\cdot3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2011}-\frac{1}{2012}\right)\)
\(S=1+2\left(\frac{1}{2}-\frac{1}{2012}\right)\)
\(S=1+2.\frac{1}{2}-2.\frac{1}{2012}\)
\(S=1+1-\frac{1}{1006}\)
\(S=\frac{2011}{1006}\)
Nho 3 tick cho mk nha
\(B=\frac{2001}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{2}{2010}+\frac{1}{2001}\)
\(B=\left(2011-1-...-1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}\)
\(B=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow\)\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}=2012\)
Vậy \(\frac{B}{A}=2012\)
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Ta có:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{2012}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{1006}\)
\(=\frac{1}{1007}+\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2012}+\frac{1}{2013}\left(1\right)\)
Mà \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow S=P\Rightarrow\left(S-P\right)^{2013}=0^{2013}=0\)
Vậy...