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CM : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
Có : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}\)\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\) đpcm
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
ta có: \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
\(\Rightarrow1-\frac{1}{1+2+3+...+n}=1-1:\frac{n.\left(n+1\right)}{2}=1-\frac{2}{n.\left(n+1\right)}\)
\(=\frac{n.\left(n+1\right)-2}{n.\left(n+1\right)}=\frac{n^2+n-2}{n.\left(n+1\right)}=\frac{\left(n+2\right).\left(n-1\right)}{n.\left(n+1\right)}\) (*)
Từ (*)
\(\Rightarrow1-\frac{1}{1+2}=\frac{4.1}{2.3};1-\frac{1}{1+2+3}=\frac{5.2}{3.4};...;1-\frac{1}{1+2+3+...+n}=\frac{\left(n+2\right).\left(n-1\right)}{n.\left(n+1\right)}\)
\(\Rightarrow E=\frac{4.1}{2.3}.\frac{5.2}{3.4}...\frac{\left(n+2\right).\left(n-1\right)}{n.\left(n+1\right)}=\frac{4.1.5.2...\left(n+1\right).\left(n-2\right).\left(n+2\right).\left(n-1\right)}{2.3.3.4....\left(n-1\right).n.n.\left(n+1\right)}\)\(=\frac{n+2}{n.n}\)
\(\Rightarrow\frac{E}{F}=E:F=\left(\frac{n+2}{n.n}\right):\frac{n+2}{n}=\frac{n+2}{n.n}.\frac{n}{n+2}=\frac{1}{n}\)
\(\Rightarrow\frac{E}{F}=\frac{1}{n}\)