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\(C=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2014}\right)}{\frac{2012}{2}+1+\frac{2011}{3}+1+......+\frac{1}{2013}+1+1}=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{..........1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+.......+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2014}\right)}{2014.\left(\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2014}\right)}=\frac{2013}{2014}\)
Ta có : 1 + 2 + 3 + ... + n = \(\frac{\left(n+1\right)n}{2}\)
Vậy nên : \(A=2013+\frac{2013}{\frac{3.2}{2}}+\frac{2013}{\frac{4.3}{2}}+...+\frac{2013}{\frac{2013.2012}{2}}\)
\(A=2013+\frac{4026}{2.3}+\frac{4016}{3.4}+...+\frac{4026}{2012.2013}\)
\(A=4026\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(A=4026\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=4026\left(1-\frac{1}{2013}\right)=4026.\frac{2012}{2013}=4024.\)
tinh chi vay