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\(A=\dfrac{1}{1\cdot2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4\cdot5}+\dfrac{1}{3\cdot4\cdot5\cdot6}+....+\dfrac{1}{9\cdot10\cdot11\cdot12}\)
\(3A=\dfrac{3}{1\cdot2\cdot3\cdot4}+\dfrac{3}{2\cdot3\cdot4\cdot5}+\dfrac{3}{3\cdot4\cdot5\cdot6}+...+\dfrac{3}{9\cdot10\cdot11\cdot12}\)
\(3A=\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4}-\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{9\cdot10\cdot11}-\dfrac{1}{10\cdot11\cdot12}\)\(3A=\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{10\cdot11\cdot12}\)
\(A=\dfrac{1}{2}-\dfrac{1}{440}\)
\(A=\dfrac{219}{440}\)
1/
A= 1/15+1/35+1/63+1/99+ ... + 1/9999
A=1/3.5+1/5.7+1/7.9+ ... +1/99.101
2A=2/3.5+2/5.7+2/7.9+ ... +2/99.101
2A=1/3-1/5+1/5-1/7+1/7-1/9+ ... + 1/99-1/101
2A=1/3-1/101
A=49/303
Sai thì thôi nhé
A= 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
A=1-1/7
A=6/7
\(A=-\frac{1}{20}+-\frac{1}{30}+-\frac{1}{42}+-\frac{1}{56}+-\frac{1}{72}+-\frac{1}{90}\)
\(\Rightarrow A=-1\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{9.10}\right)\)
\(A=-1\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(\Rightarrow A=-\frac{3}{20}\)
\(A=\frac{-1}{20}-\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(A=-\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(A=\frac{-3}{20}\)
#
Bài 1 :
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(S=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)
Bài 2 :
\(S=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)
\(S=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)
Bài 3 :
\(3S=\frac{3}{4\times7}+\frac{3}{7\times11}+...+\frac{3}{19\times22}\)
\(3S=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{22}\)
\(3S=\frac{1}{4}-\frac{1}{22}\)
\(S=\frac{18}{88}\div3=\frac{6}{88}\)
Bài 1:
Ta thấy:
\(\frac{1}{2}>\frac{1}{6};\frac{1}{3}>\frac{1}{6};\frac{1}{4}>\frac{1}{6};\frac{1}{5}>\frac{1}{6};\frac{1}{6}=\frac{1}{6}\)
\(=>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}>\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\)
\(=>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}>\frac{5}{6}\)
Bài 2:
Đặt \(A=\frac{1}{5}+\frac{1}{45}+\frac{1}{117}+...+\frac{1}{1517}\)
Ta thấy \(\frac{1}{5}=\frac{1}{1.5};\frac{1}{45}=\frac{1}{5.9};\frac{1}{117}=\frac{1}{9.13}\)
Theo quy luật như vậy ta có các số tiếp theo là:
\(\frac{1}{13.17}=\frac{1}{221};\frac{1}{17.21}=\frac{1}{357};\frac{1}{21.25}=\frac{1}{525};\frac{1}{25.29}=\frac{1}{725};...\)
Ta có \(A=\frac{1}{5}+\frac{1}{45}+\frac{1}{117}+...+\frac{1}{1517}\)
\(=>A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{27.31}\)
\(=>4A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{27.31}\)
\(=>4A=\frac{5-1}{1.5}+\frac{9-5}{5.9}+\frac{13-9}{9.13}+...+\frac{31-27}{27.31}\)
\(=>4A=\frac{5}{1.5}-\frac{1}{1.5}+\frac{9}{5.9}-\frac{5}{5.9}+\frac{13}{9.13}-\frac{9}{9.13}+...+\frac{31}{27.31}-\frac{27}{27.31}\)
\(=>4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{27}-\frac{1}{31}\)
\(=>4A=1-\frac{1}{31}=\frac{30}{31}=>A=\frac{30}{31}.\frac{1}{4}=\frac{15}{62}\)
