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Xét \(\frac{n}{1+n^2+n^4}=\frac{n}{n^4+2n^2+1-n^2}=\frac{n}{\left(n^2+1\right)^2-n^2}=\frac{n}{\left(n^2-n+1\right)\left(n^2+n+1\right)}=\frac{1}{2}\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right)\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{2013^2-2013+1}-\frac{1}{2013^2+2013+1}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{2013^2+2013+1}\right)=...\)
\(S=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{1}{2014}}\)
Xét mẫu:
\(\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{1}{2014}\)
= \(\left(1+\frac{2013}{2}\right)+\left(1+\frac{2012}{3}\right)+...+\left(1+\frac{1}{2014}\right)+1\)
= \(\frac{2014}{2}+\frac{2014}{3}+....+\frac{2014}{2013}+\frac{2014}{2014}\)
= \(2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(\Rightarrow S=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}\)
\(\Rightarrow S=\frac{1}{2014}\)
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Xét Tử số của A ta có:
\(2014+\frac{2013}{2}+\frac{2012}{3}+....+\frac{2}{2013}=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(\frac{1}{2014}+1\right)\)\(TS=\frac{2015}{2}+\frac{2015}{3}+....+\frac{2015}{2014}+\frac{2015}{2015}\)
\(TS=2015.\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)\)
\(A=\frac{2015.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)}=2015\)
Đặt \(S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+....+\frac{2013}{1+2013^2+2013^4}\)
Xét:
\(\frac{k}{k+k^2+k^4}=\frac{1}{2}\cdot\frac{k^2+k+1-k^2+k-1}{k^4+k^2+1}\)
\(=\frac{1}{2}\cdot\frac{k\left(k+1\right)+1-k\left(k-1\right)-1}{\left(k^2+1\right)^2-k^2}\)
\(=\frac{1}{2}\left[\frac{1}{k\left(k-1\right)+1}-\frac{1}{k\left(k+1\right)+1}\right]\)
Áp dụng :
\(S=\frac{1}{2}\left[\frac{1}{1\cdot0+1}-\frac{1}{1\cdot2+1}+\frac{1}{2\cdot1+1}-\frac{1}{2\cdot3+1}+.....+\frac{1}{2013\cdot2012+1}-\frac{1}{2013\cdot2014+1}\right]\)
\(=\frac{2027091}{4054183}\)