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Với \(k\in N;k>0\) Ta có :
\(\frac{1}{k\left(k+1\right)\left(k+2\right)}=\frac{1}{2}.\frac{\left(k+2\right)-k}{k\left(k+1\right)\left(k+2\right)}=\frac{1}{2}\left(\frac{1}{k\left(k+1\right)}-\frac{1}{\left(k+1\right)\left(k+2\right)}\right)\)
Áp dụng ta có :
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.....+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{n\left(n+1\right)}\right)=\frac{1}{2}.\frac{n\left(n+1\right)-2}{2n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{4n\left(n+1\right)}\)(đpcm)
Ta có :
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{4n\left(n+1\right)}\)
\(\Leftrightarrow\)\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{\left(n-1\right)n\left(n+1\right)}=\frac{2\left(n-1\right)\left(n+2\right)}{4n\left(n+1\right)}\)
\(\Leftrightarrow\)\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}=\frac{n\left(n-1\right)+2\left(n-1\right)}{2n\left(n+1\right)}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{n\left(n+1\right)}=\frac{n^2-n+2n-2}{2n^2+2n}\)
\(\Leftrightarrow\)\(\frac{n\left(n+1\right)}{2n\left(n+1\right)}-\frac{2}{2n\left(n+1\right)}=\frac{n^2+n-2}{2n^2+2n}\)
\(\Leftrightarrow\)\(\frac{n^2+n-2}{2n^2+2n}=\frac{n^2+n-2}{2n^2+2n}\) với \(n\ge2\)
Vậy ...
Ta có :
\(1-\frac{3}{n\left(n+2\right)}=\frac{n^2+2n-3}{n\left(n+2\right)}=\frac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(\Rightarrow A=\frac{1.5}{2.4}.\frac{2.6}{3.5}...\frac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(=\left(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n-1}{n}\right)\left(\frac{5}{4}.\frac{6}{5}.\frac{7}{6}...\frac{n+3}{n+2}\right)\)
\(=\frac{1}{n}.\frac{n+3}{4}=\frac{n+3}{n}.\frac{1}{4}\ge\frac{1}{4}\left(dpcm\right)\)
\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)
\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(=50.\frac{9}{50}=9\)
