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\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(\Rightarrow T=\frac{1004}{1005}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\frac{2010}{2011}\)
\(\Rightarrow A=\frac{1005}{2011}\)
\(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)
\(\Rightarrow A=4\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{2008.2010}\right)\)
\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\right)\right]\)
\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2010}\right)\right]\Rightarrow A=4\left(\frac{1}{2}.\frac{502}{1005}\right)\Rightarrow A=4.\frac{251}{1005}\Rightarrow A=\frac{1004}{1005}\)
\(B=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\)
\(\Rightarrow B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+....+\frac{1}{30.33}\)
\(\Rightarrow B=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+.....+\frac{1}{30}-\frac{1}{33}\right)\)
\(\Rightarrow B=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\Rightarrow B=\frac{1}{3}.\frac{10}{33}\Rightarrow B=\frac{10}{99}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(B=1-\frac{1}{101}=\frac{100}{101}\)
\(C=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)
\(C=3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)
\(C=3\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\right)\)
\(C=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(C=3\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{27}{20}\)
\(D=\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
\(D=\frac{7}{2}B=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
a,\(\frac{2}{1.3}+...\frac{2}{99.101}\)
\(=\frac{3-1}{1.3}+...+\frac{101-99}{99.101}\)
\(=\frac{3}{1.3}-\frac{1}{1.3}+...+\frac{101}{99.101}-\frac{99}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(\frac{100}{101}\)
a, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
=2.(\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\))
=\(2.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
=\(\frac{2}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{100}{101}\)
b, \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
=\(5.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)
=\(5.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
=\(\frac{250}{101}\)
\(=\frac{5}{2}.\frac{100}{101}\)
a,21.321.3+23.523.5+25.725.7+....+299.101
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)
=>\(\frac{1}{1}-\frac{1}{101}\)
=>\(\frac{100}{101}\)
b,
51.351.3+53.553.5+55.755.7+....+599.101
=>\(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{99.101}\right)\)
=>\(\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)
=>\(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
=>\(\frac{5}{2}.\frac{100}{101}\)
=>\(\frac{250}{101}\)
A = \(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
=\(7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{70}\right)\)
=\(7.\frac{3}{35}\)
=\(\frac{3}{5}\)
B=\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
=\(\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)
=\(\frac{1}{2}.\frac{2}{75}\)
=\(\frac{1}{75}\)
1)\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
2)\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2\times\frac{502}{1005}\)
\(=\frac{1004}{1005}\)
tự làm tiếp nhé
1.= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
= \(1-\frac{1}{101}\) = \(\frac{100}{101}\)
2.= \(2\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)
= \(2\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
= \(2\cdot\left(\frac{1}{2}-\frac{1}{2010}\right)\) = \(2\cdot\frac{502}{1005}\) = \(\frac{1004}{1005}\)