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Ta có
\(\frac{1}{1.6}+\frac{1}{6.11}+......+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)
\(=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+.....+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)
\(=\frac{1}{5}\left(1-\frac{1}{5n+6}\right)\)
\(=\frac{1}{5}.\left[\frac{\left(5n+6\right)-1}{\left(5n+6\right)}\right]\)
\(=\frac{1}{5}.\frac{5n+5}{5n+6}\)
\(=\frac{n+1}{5n+6}\)
\(\Rightarrow\frac{1}{1.6}+\frac{1}{6.11}+......+\frac{1}{\left(5n+1\right)\left(5n+6\right)}=\frac{n+1}{5n+6}\) ( đpcm )
Đặt A = \(\frac{1}{1.6}+\frac{1}{6.11}+..+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)
5A = \(\frac{5}{1.6}+\frac{5}{6.11}+..+\frac{5}{\left(5n+1\right)\left(5n+6\right)}\)
= \(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+..+\frac{1}{5n+1}-\frac{1}{5n+6}\)
= \(\frac{1}{1}-\frac{1}{5n+6}=\frac{5n+6-1}{5n+6}=\frac{5n+5}{5n+6}=\frac{5\left(n+1\right)}{5n+6}\)
=> A = \(=\frac{5\left(n+1\right)}{5n+6}:5=\frac{5\left(n+1\right)}{5n+6}\cdot\frac{1}{5}=\frac{n+1}{5n+6}\)
VẬy VT = VP ĐT Đ CM
Ta có: \(\frac{3}{9.14}+\frac{3}{14.19}+...+\frac{3}{\left(5n-1\right)\left(5n+4\right)}\)
\(=\frac{3}{5}\left(\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{\left(5n-1\right)\left(5n+4\right)}\right)\)
\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{5n-1}-\frac{1}{5n+4}\right)\)
\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{5n-1}\right)\)
\(=\frac{1}{15}-\frac{3}{5\left(5n-1\right)}\)
Vì \(\frac{1}{15}-\frac{3}{5\left(5n-1\right)}< \frac{1}{15}\) nên \(\frac{3}{9.14}+\frac{3}{19.19}+...+\frac{3}{\left(5n-1\right)\left(5n+4\right)}< \frac{1}{15}\left(đpcm\right)\)
Đặt A= \(\frac{3}{9.14}+\frac{3}{14.19}+...+\frac{3}{\left(5n+1\right).\left(5n+4\right)}\)
\(\Rightarrow A=3.\left(\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{\left(5n-1\right)\left(5n+4\right)}\right)\)
\(=3.5.\frac{1}{5}.\left(\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{\left(5n-1\right)\left(5n+4\right)}\right)\)
\(=\frac{3}{5}\left(\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{\left(5n-1\right)\left(5n+4\right)}\right)\)
\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{5n-1}-\frac{1}{5n+4}\right)\)
\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{5n+4}\right)\)
\(\Rightarrow\)\(A< \frac{3}{5}.\frac{1}{9}\)\(\Rightarrow A< \frac{1}{15}\)(đpcm)
đpcm<=> 5/9.14+5/14.19+...+5/(5n-1)(5n+4)<1/9
<=>1/9-1/5n+4<1/9
<=>5n-5/45n+36<1/9(đúng với mọi n>=2)
Vậy ddpcm là đúng
Ta có:\(\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)
\(=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)
\(=\frac{1}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)
\(=\frac{1}{5}.\left(1-\frac{1}{5n+6}\right)\)
\(=\frac{1}{5}.\left(\frac{5n+5}{5n+6}\right)=\frac{n+1}{5n+6}\left(\text{đ}pcm\right)\)