Tính
a) K =\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2015.2016.2017}\)
b) O = \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{2015.2016.2017.2018}\)
K
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16 tháng 1 2017
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}-3x=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)-3x=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)-3x=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)
\(\Leftrightarrow\frac{4949}{19800}-3x=\frac{451}{8120}\)
\(\Leftrightarrow x\approx0,0648\)
7 tháng 3 2016
a)\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+....+\(\frac{1}{100.101}\)=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+....+\(\frac{1}{100}\)-\(\frac{1}{101}\)=1-\(\frac{1}{101}\)=\(\frac{100}{101}\)
b)\(\frac{1}{1.2.3}\)+\(\frac{1}{2.3.4}\)+....+\(\frac{1}{28.29.30}\)=\(\frac{868}{3480}\)=\(\frac{217}{870}\)
c)\(\frac{1}{1.2.3.4}\)+\(\frac{1}{2.3.4.5}\)+....+\(\frac{1}{27.28.29.30}\)=\(\frac{24354}{438480}\)=\(\frac{451}{8120}\)
14 tháng 1 2017
a) Đặt A=\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+.....+\frac{1}{98\cdot99\cdot100}\)
\(\Rightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+....+\frac{2}{98\cdot99\cdot100}\)
\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+.....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)
2A=\(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}=\frac{4949}{9900}\) =>A=\(\frac{4949}{9900}\div2=\frac{4949}{19800}\)
Đặt B=\(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29\cdot30}\)
=>3B=\(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+....+\frac{3}{27\cdot28\cdot29\cdot30}\)
3B=\(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+.....+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\)
3B=\(\frac{1}{1\cdot2\cdot3}-\frac{1}{28\cdot29\cdot30}=\frac{1353}{8120}\)
=>B=\(\frac{1353}{8120}\div3=\frac{451}{8120}\)
Ta có : A-3x=B=>3x=A-B=\(\frac{4949}{19800}\)-\(\frac{451}{8120}\)\(\approx\frac{1}{5}\)=>x=\(\frac{1}{5}\div3\)=\(\frac{1}{15}\)
KK
14 tháng 1 2017
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}-3x=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)-3x=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)-3x=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)
\(\Rightarrow\frac{4949}{19800}-3x=\frac{451}{8120}\)
\(\Rightarrow3x=\frac{4949}{19800}-\frac{451}{8120}\)
\(\Rightarrow x=\left(\frac{4949}{19800}-\frac{451}{8120}\right):3\)