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$\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}$4n(n+2)(n+4) =n+4−nn(n+2)(n+4) =1n(n+2) −1(n+2)(n+4) $\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}<\frac{1}{3}$B9 =11.3 −13.5 +13.5 −15.7 +...+125.27 −127.29 =13 −127.29 <13 $\Rightarrow B<3$
\(\frac{36}{1\cdot3\cdot5}+\frac{36}{3\cdot5\cdot7}+\frac{36}{5\cdot7\cdot9}+...+\frac{36}{25\cdot27\cdot29}\)
\(=9\left[\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{25\cdot27\cdot29}\right]\)
\(=9\left[\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\right]\)
\(=9\left[\frac{1}{3}-\frac{1}{783}\right]=9\cdot\frac{260}{783}=\frac{260}{87}\)
Đặt \(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{1}{1.3}-\frac{1}{27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{261}{783}-\frac{1}{783}\)
\(\Rightarrow\frac{1}{9}A=\frac{260}{783}\)
\(\Rightarrow A=\frac{260}{783}\div\frac{1}{9}\)
\(\Rightarrow A=\frac{2340}{783}=\frac{260}{87}\)
a)\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
=\(\frac{9.4}{1.3.5}+\frac{9.4}{3.5.7}+\frac{9.4}{5.7.9}+...+\frac{9.4}{25.27.29}\)
=\(9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
=\(9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
=\(9.\left(\frac{1}{3}-\frac{1}{27.29}\right)=9.\left(\frac{1}{3}-\frac{1}{783}\right)=9.\left(\frac{261}{783}-\frac{1}{783}\right)=9.\frac{260}{783}\)
=\(\frac{260}{87}\)
b)
ta có: \(3=\frac{261}{87}>\frac{260}{87}\)
vậy A<3
\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+.....+\frac{36}{25.27.29}\)
\(=9\left(\frac{4}{3.\left(1.5\right)}+\frac{4}{5\left(3.7\right)}+\frac{4}{7.\left(5.9\right)}+.....+\frac{4}{27\left(25.29\right)}\right)\)
\(=9\left(\frac{1}{3}\left(1-\frac{1}{5}\right)+\frac{1}{5}\left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{7}\left(\frac{1}{5}-\frac{1}{9}\right)+...+\frac{1}{27}\left(\frac{1}{25}-\frac{1}{29}\right)\right)\)
\(=9\left(\frac{1}{3}-\frac{1}{3.5}+\frac{1}{5.3}-\frac{1}{5.7}+\frac{1}{7.5}-\frac{1}{7.9}+...+\frac{1}{27.25}-\frac{1}{27.29}\right)\)
\(=9\left(\frac{1}{3}-\frac{1}{27.29}\right)< 9.\frac{1}{3}=3\)
Vậy A < 3.
\(A=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{25.27.29}\right)\)
\(A=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(A=9.\left(\frac{1}{3}-\frac{1}{783}\right)\)
\(A=9.\frac{260}{87}=\frac{260}{87}
\(\frac{C}{9}=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{21.23.25}+\frac{4}{23.25.27}.\)
\(\frac{C}{9}=\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+\frac{9-5}{5.7.9}+...+\frac{25-21}{21.23.25}+\frac{27-23}{23.25.27}\)
\(\frac{C}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{21.23}-\frac{1}{23.25}+\frac{1}{23.25}-\frac{1}{25.27}\)
\(\frac{C}{9}=\frac{1}{3}-\frac{1}{25.27}\Rightarrow C=\frac{9\left(25.9-1\right)}{25.27}=\frac{25.9-1}{25.3}=3-\frac{1}{25.3}< 3\)
Áp dụng: \(\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}\)
\(\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}