A = \(\frac{1}{1x3}\)+ \(\frac{1}{3x5}\)+ \(\frac{1}{5x7}\)+ ..... + \(\frac{1}{97x99}\)
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Tôi đặt S cho nhanh, đừng hỏi tại sao còn bạn chứ là A nhé :))
\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow M=1-\frac{1}{100}\)
\(\Rightarrow M=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
\(b,N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\Rightarrow N=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}\)
\(\Rightarrow N=\frac{1.98}{2.99}=\frac{49.2}{2.99}=\frac{49}{99}\)
\(a,M=1-\frac{1}{100}=\frac{99}{100}\)
\(b=2N=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{97x99}\)
\(=1-\frac{1}{99}=\frac{98}{99}\)
=>\(N=\frac{98}{99}:2=\frac{49}{99}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{9.11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\frac{10}{11}\)
\(=\frac{5}{11}\)
\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{9\times11}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)
\(=\frac{1}{2}\times\frac{10}{11}\)
\(=\frac{5}{11}\)
\(S.2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(S.2=\frac{1}{1}-\frac{1}{11}\)
\(S.2=\frac{10}{11}\)
\(S=\frac{10}{11}:2\)
\(S=\frac{5}{11}\)
\(2A=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{999x1001}\)
\(2A=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{1001-999}{999x1001}\)
\(2A=\frac{3}{1x3}-\frac{1}{1x3}+\frac{5}{3x5}-\frac{3}{3x5}+\frac{7}{5x7}-\frac{5}{5x7}+...+\frac{1001}{999x1001}-\frac{999}{999x1001}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\)
\(2A=1-\frac{1}{1001}=\frac{1000}{1001}\)=> A = 500/1001
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+\frac{1}{5}-\frac{1}{7}-\frac{1}{6}+\frac{1}{8}+\frac{1}{7}-\frac{1}{9}-\frac{1}{8}+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}.\frac{11}{10}\)
\(\Rightarrow S=\frac{11}{20}\)
\(=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{2013x2015}\)
\(=\frac{1}{2}x\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{2013x2015}\right)\)
\(=\frac{1}{2}x\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}x\left(1-\frac{1}{2015}\right)\)
\(=\frac{1}{2}x\frac{2014}{2015}\)
\(=\frac{1007}{2015}\)
1/1-1/3+1/3-1/5+1/5-1/7+....+1/2013-1/2015
=1/1-1/2015
=2014/2015
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{13.15}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
\(=\frac{14}{15}\)
98/99
ghi rõ lời giải ra nha