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Đặt tổng trên = A
Có : A = 1/1.2.3 + 1/2.3.4 + ...... + 1/9.10.11
2A = 2/1.2.3 + 2/2.3.4 + ...... + 2/9.10.11
= 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ....... + 1/9.10 - 1/10.11
= 1/1.2 - 1/10.11
= 1/2 - 1/110 = 27/55
=> A = 27/55 : 2 = 27/110
Tk mk nha
B = 2 - 4 - 6 + 8 + 10 - 12 -14 + 16 + ...+ 2010 - 2012 - 2014 + 2016
= (2 - 4 - 6 + 8 ) + ( 10 - 12 -14 +16 ) + ...+ ( 2010 - 2012 - 2014 + 2016 )
= 0 + 0 +...+ 0 + 0 (có 252 số hạng 0)
= 0
Ta có: Từ 2 đến 2016 có \(\frac{2016-2}{2}+1=1008\)
B=(2-4)-(6-8)+(10-12)-(14-16)+...+(2010-2012)-(2014-2016) như vậy có 1008:2=504 nhóm
=-2+2-2+2+...-2+2 có 504 số hạng trong đó có: 525 số -2 và 525 số +2
=0
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{18\cdot19\cdot20}\)
\(A=\frac{1}{2}\cdot\frac{2}{1\cdot2\cdot3}+\frac{1}{2}\cdot\frac{2}{2\cdot3\cdot4}+\frac{1}{2}\cdot\frac{2}{3\cdot4\cdot5}+...+\frac{1}{2}\cdot\frac{2}{18\cdot19\cdot20}\)
\(A=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{18\cdot19\cdot20}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2.3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-0-0-...-0-\frac{1}{380}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)\)
\(A=\frac{1}{2}\cdot\frac{189}{380}\)
\(A=\frac{189}{760}\)
\(N=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\)
=\(\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)
=\(\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{10}{33}\)
\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4970}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{70.71}\)
\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{70}-\frac{1}{71}\)
\(M=1-\frac{1}{71}\)
\(M=\frac{70}{71}\)
\(N=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(N=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(N=\frac{1}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\right)\)
\(N=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(N=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(N=\frac{1}{3}.\frac{10}{33}\)
\(N=\frac{10}{99}\)
\(A=\left(\frac{-2}{3}+1\frac{1}{4}-\frac{1}{6}\right).\frac{-12}{5}\)
=\(\left(\frac{-8+15-2}{12}\right).\frac{-12}{5}\)\(=\frac{5}{12}.\frac{-12}{5}=-1\)
Làm lại câu a
\(2S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(2S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(2S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(2S=1-\frac{1}{100}\)suy ra \(2S=\frac{99}{100}\)
\(S=\frac{99}{100}:2\)suy ra \(S=\frac{99}{200}\)
a, 2S=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(2S=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{100}\)
\(2S=1-\frac{1}{100}\)suy ra \(2S=\frac{99}{100}\)
\(S=\frac{99}{100}:2=\frac{99}{200}\)
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(=\frac{1}{3}.\frac{10}{33}\)
\(=\frac{10}{99}\)
Đúng không Bạch Dương ?
Ta có: \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{2.9}+\frac{1}{6.9}+\frac{1}{12.9}+...+\frac{1}{110.9}\)
\(=\frac{1}{9}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
\(=\frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(=\frac{1}{9}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=\frac{1}{9}\left(\frac{1}{1}-\frac{1}{11}\right)\)
\(=\frac{1}{9}.\frac{10}{11}\)
\(=\frac{10}{99}\)
Vậy \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}=\frac{10}{99}\)
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