\(.S=3.\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{96.101}\right)\)

\(\Rightarrow S=3.\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+...+\frac{1}{96}-\frac{1}{101}\right)\)

\(\Rightarrow S=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)\)

\(\Rightarrow S=\frac{3}{5}.\left(\frac{100}{101}\right)\)

\(S=\frac{60}{101}\)

\(\frac{100}{101}\)nha

bạn tự tính

tíc mình nha

S=3/1.6+3/6.11+3/11.16+...+3/96.101

=>S=1/1.6+1/6.11+1/11.16+...+1/96.101

S=1-1/6+1/6-1/11+1/11-1/16+...+1/96-1/101

S=1-1/101

S=100/101

\(\frac{3}{1.6}+\frac{3}{6.11}+\frac{3}{11.16}+...+\frac{3}{96.101}\)

\(=3.\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)\)

\(=\frac{3}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)

\(=\frac{3}{5}.\left(1-\frac{1}{101}\right)\)

\(=\frac{3}{5}.\frac{100}{101}\)

\(=\frac{60}{101}\)

đm dễ thế này thi tự làm đi hỏi cc

\(A=\frac{1}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{5}{96\cdot101}\right)\)

\(A=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)

\(A=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{101}\right)\)

\(A=\frac{1}{5}\cdot\frac{100}{101}\)

\(A=\frac{20}{101}\)

A = 1/5(1-1/6+1/6-1/11+1/11-1/16+.....+1/96-1/101)

    = 1/5(1-1/101)=20/101

ta có : 1/1.6+1/6.11+1/11.16+....+1/96.101

= 1/5.5/1.6+ 1/5.5/6.11+1/5.5/11.16+...+1/5.5/96.101

=1/5 . ( 5/1.6+5/6.11+5/11.16+...+5/96.101)

=1/5 . ( 1/1-1/6 +1/6-1/11+1/11-1/16+....+1/96-1/101)

=1/5 . (1/1-1/101)

=1/5 . 100/101

= 20/101

5A=\( 1-{1\over 6}+{1\over 6}-{1\over 11}+...{1\over 96}-{1\over 101}\)

  =\(1- {1 \over 101}={100 \over 101}\)

suy ra A =\({20 \over 101}\)

\(A=\frac{1}{5}\cdot\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+\frac{...5}{96\cdot101}\right)\)

\(A=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)

\(A=\frac{1}{5}\left(\frac{1}{1}+0+0+0+...+0-\frac{1}{101}\right)\)

\(A=\frac{1}{5}\left(1-\frac{1}{101}\right)\)

\(A=\frac{1}{5}\cdot\frac{100}{101}\)

\(A=\frac{20}{101}\)

a,1/1-1/4+1/4-1/7+...+1/2008-1/2011

=(1-1/2011)+(-1/4+1/4)+...+(-1/2008+1/2008)

=1-1/2011+0+...+0

=1-1/2011

=2010/2011

\(C=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)

\(C=\frac{1}{5}\left(1-\frac{1}{101}\right)\)

\(C=\frac{1}{5}.\frac{100}{101}=\frac{20}{101}\)

\(5C=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\)

\(5C=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\)

\(5C=1-\frac{1}{101}\)

\(C=\frac{100}{\frac{101}{5}}\)

bổ sung 20/101

\(b\)) \(Q=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)

\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)

\(a\)) Mình giải theo cách khác:

Chú ý rằng : \(\frac{3}{2.5}=\frac{1}{2}-\frac{1}{5};\frac{3}{5.8}=\frac{1}{5}-\frac{1}{8};\frac{3}{8.11}=\frac{1}{8}-\frac{1}{11};...;\frac{3}{17.20}=\frac{1}{17}-\frac{1}{20}\)

Do đó: \(P=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)