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S = \(\frac{\left(1+90\right)\times90\div2}{4}\)
S = \(\frac{4095}{4}\)
4/42+4/56+4/72+4/90+4/110
=4*(1/42+1/56+1/72+1/90+1/110)
=4*(1/(6*7)+1/(7*8)+1/(8*9)+1/(9*10)+1/(10*11))
=4*(1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11)
=4*(1/6-1/11)
=4*5/66
=10/33
\(\frac{4}{6\cdot7}+\frac{4}{7\cdot8}+\frac{4}{8\cdot9}+\frac{4}{9\cdot10}+\frac{4}{10\cdot11}\)
\(4\cdot\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\right)\)
\(4\cdot\frac{5}{66}\)
\(\frac{10}{33}\)
\(A=\frac{4}{2}+\frac{4}{6}+\frac{4}{12}+\frac{4}{20}+\frac{4}{30}+\frac{4}{42}\)
\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+\frac{4}{4.5}+\frac{4}{5.6}+\frac{4}{6.7}\)
\(A=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=4\left(1-\frac{1}{7}\right)\)
\(A=4.\frac{6}{7}\)
\(A=\frac{24}{7}\)
\(A=\frac{4}{2}+\frac{4}{6}+\frac{4}{12}+\frac{4}{20}+\frac{4}{30}+\frac{4}{42}=4\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=4\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=4\left(1-\frac{1}{7}\right)=\frac{6}{7}.4=\frac{24}{7}\)
M = \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2015.2017}\)4/1.3 + 4/3.5 + 4/5.7 + ... + 4/2015.2017
M = \(2.\frac{2}{1.3}+2.\frac{2}{3.5}+2.\frac{2}{5.7}+...+2.\frac{2}{2015.2017}\) 2 . 2/1.3 + 2 . 2/3.5 + 2 . 2/5.7 + ... + 2 . 2/2015.2017
M = 2 . ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2015.2017 )
M = 2 . ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2015 - 1/2017 )
M = 2 . ( 1 - 1/2017 )
M = 2 . 2016/2017
M = 4032/2017
\(M=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(M=2\left(1-\frac{1}{2017}\right)\)
\(M=\frac{4032}{2017}\)
\(S=\frac{1}{4}+\frac{2}{4}+\frac{3}{4}+....+\frac{90}{4}\)
\(=\frac{1+2+3+...+90}{4}\)
\(=\frac{\frac{90\left(90+1\right)}{2}}{4}\)
\(=\frac{4095}{4}\)
Ta thấy \(\frac{1}{3}-\frac{1}{7}=\frac{7-3}{3.7}=\frac{4}{3.7}\)
\(\frac{1}{7}-\frac{1}{11}=\frac{11-7}{7.11}=\frac{4}{7.11}\)
..........................
\(\frac{1}{1023}-\frac{1}{1027}=\frac{1027-1023}{1023.1027}=\frac{4}{1023.1027}\)
=> \(\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{1023.1027}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{1023}-\frac{1}{1027}\)
=> =\(\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3.1027}\)
Ta có: \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{1023.1027}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{1023}-\frac{1}{1027}\)
\(=\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3081}\)
\(\frac{1}{3}-\frac{1}{2027}\)