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\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A=1-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
a)A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{2009.2010}\)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2009}-\frac{1}{2010}\)
A=1-\(\frac{1}{2010}\)=\(\frac{2009}{2010}\)
c)C=\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+......+\frac{1}{2006.2008}\)
C=\(\frac{1}{2}\).(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+..+\frac{1}{2006}-\frac{1}{2008}\))
C=\(\frac{1}{2}\).(\(\frac{1}{2}-\frac{1}{2008}\))
C=\(\frac{1}{2}\).\(\frac{1003}{2008}\)=\(\frac{1003}{4016}\)
Câu b mình chưa nghĩ ra
Chúc bạn học tốt!
a) A = \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + ...+ \(\frac{1}{2009.2000}\)
= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + ... + \(\frac{1}{2009}\) - \(\frac{1}{2000}\)
= 1 - \(\frac{1}{2000}\) = \(\frac{1999}{2000}\)
b) B = \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + \(\frac{1}{3.4.5}\) + ... + \(\frac{1}{1998.1999.2000}\)
= \(\frac{1}{2}\) ( \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + \(\frac{2}{3.4.5}\) + ... + \(\frac{2}{1998.1999.2000}\))
= \(\frac{1}{2}\) (\(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + \(\frac{1}{3.4}\) - \(\frac{1}{4.5}\) + ... + \(\frac{1}{1998.1999}\) - \(\frac{1}{1999.2000}\))
= \(\frac{1}{2}\) (\(\frac{1}{1.2}\) - \(\frac{1}{1999.2000}\))
= \(\frac{1}{2}\) (\(\frac{1}{2}\) - \(\frac{1}{3998000}\))
= \(\frac{1}{4}\) - \(\frac{1}{7996000}\) = ?
c) C = \(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + \(\frac{1}{6.8}\) + ... + \(\frac{1}{2006.2008}\)
= \(\frac{1}{2}\) (\(\frac{1}{2}\) - \(\frac{1}{4}\)) + \(\frac{1}{2}\)(\(\frac{1}{4}\) - \(\frac{1}{6}\)) + ... + \(\frac{1}{2}\)(\(\frac{1}{2006}\) - \(\frac{1}{2008}\))
= \(\frac{1}{2}\)(\(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{6}\) + ... + \(\frac{1}{2006}\) - \(\frac{1}{2008}\))
= \(\frac{1}{2}\)(\(\frac{1}{2}\) - \(\frac{1}{2008}\))
= \(\frac{1}{2}\) . \(\frac{1003}{2008}\) = \(\frac{1003}{4016}\).
\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)x=\frac{1}{3}\left(2014.2015.2016-2013.2014.2015........+2.3.4-1.2.3+1.2.3-0.1.2\right)\)
\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)
\(x=\frac{1}{3.2029104}.2014^2.2015^2.2016=\)
\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)
Làm bậy, mà đúng
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ … + \(\frac{1}{99.100}\)
= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)-\(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ … + \(\frac{1}{99}\)- \(\frac{1}{100}\)
= \(\frac{1}{1}\)- \(\frac{1}{100}\)
= \(\frac{99}{100}\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
a)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+....+\frac{1}{98\cdot99\cdot100}\)
\(=\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+\frac{5-3}{3\cdot4\cdot4}+....+\frac{100-98}{98\cdot99\cdot100}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)