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\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2017.2019}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{2017.2019+1}{2017.2019}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{2018^2}{2017.2019}\)
\(=\frac{2}{1}.\frac{2018}{2019}=\frac{4036}{2019}\)
\(S=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{7.9}+\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\right)\)
Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\)
2A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{7.9}\)
2A = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\)
2A = \(1-\frac{1}{9}=\frac{8}{9}\)
A = \(\frac{8}{9}:2=\frac{4}{9}\)
Đặt B = \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\)
2B = \(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{8.10}\)
2B = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)
2B = \(\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
B = \(\frac{2}{5}:2=\frac{1}{5}\)
Thay A và B vào S ta được:
\(S=\frac{4}{9}+\frac{1}{5}=\frac{29}{45}\)
\(S=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{7.9}+\frac{1}{8.10}\)
\(\Rightarrow S=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}\)
\(S=\frac{1}{2}\left(\frac{8}{9}+\frac{2}{5}\right)\)
\(S=\frac{1}{2}.\frac{58}{45}\)
\(S=\frac{29}{45}\)
\(S=\frac{1}{1.2}+\frac{1}{3.4}+.........+\frac{1}{199.200}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(=\frac{4}{9}-\frac{1}{5}\)
\(=\frac{11}{45}\)
Có \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)..........\)\(\left(1+\frac{1}{2014.2016}\right)\)
=\(\left(\frac{1.3}{1.3}+\frac{1}{1.3}\right)\left(\frac{2.4}{2.4}+\frac{1}{2.4}\right)....\left(\frac{2014.2016}{2014.2016}+\frac{1}{2014.2016}\right)\)
=\(\left(\frac{2^2-1}{1.3}+\frac{1}{2.4}\right)\left(\frac{3^2-1}{2.4}+\frac{1}{2.4}\right)......\left(\frac{2015^2-1}{2014.2016}+\frac{1}{2014.2016}\right)\)
=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}......\frac{2015.2015}{2014.2016}\)
=\(\frac{2.2.3.3.....2015.2015}{1.3.2.4....2014.2015}\)
=\(\frac{\left(2.3...2015\right).\left(2.3.....2015\right)}{\left(1.2....2014\right).\left(3.4.....2016\right)}=\frac{2015.2}{2016}=\frac{4030}{2016}\)
Ta có:
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+\frac{1}{5.7}+\frac{1}{6.8}+\frac{1}{7.9}+\frac{1}{8.10}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}....+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}=\frac{1}{2}.\left(\frac{8}{9}+\frac{2}{5}\right)=\frac{1}{2}.\frac{58}{45}=\frac{29}{45}\)
= 4/1.3 x 9/2.4 x 16/3.5 x...x 10000/99.101
= 2.2/1.3 x 3.3/2.4 x 4.4/3.5 x..x 100.100/99.101
= (2.3.4. ... 100/1.2.3. .... 99) x (2.3.4. ... .100/3.4.5. ... .101)
= 100.2/101
=200/101
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(\Rightarrow A=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{99.101+1}{99.101}\)
\(\Rightarrow A=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{10000}{99.101}\)
\(\Rightarrow A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)
\(\Rightarrow A=\frac{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}\)
\(\Rightarrow A=\frac{100.2}{101}=\frac{200}{101}\)
