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\(\Leftrightarrow B=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(\Leftrightarrow B=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(\Leftrightarrow B=\frac{3}{5}.\frac{100}{101}\)
\(\Leftrightarrow B=\frac{60}{101}\)
\(.S=3.\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{96.101}\right)\)
\(\Rightarrow S=3.\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(\Rightarrow S=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(\Rightarrow S=\frac{3}{5}.\left(\frac{100}{101}\right)\)
\(S=\frac{60}{101}\)
\(\frac{3}{1.6}+\frac{3}{6.11}+\frac{3}{11.16}+...+\frac{3}{96.101}\)
\(=3.\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)\)
\(=\frac{3}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(=\frac{3}{5}.\left(1-\frac{1}{101}\right)\)
\(=\frac{3}{5}.\frac{100}{101}\)
\(=\frac{60}{101}\)
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
\(b\)) \(Q=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)
\(a\)) 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: \(P=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
Ta có :
\(S=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}\)
\(S=5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}\right)\)
\(S=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}\right)\)
\(S=5\left(1-\frac{1}{26}\right)\)
\(S=5.\frac{25}{26}\)
\(S=\frac{125}{26}\)
Vậy \(S=\frac{125}{26}\)
Chúc bạn học tốt ~
\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{46.51}\)
\(=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{46.51}\right)\)
\(=\frac{1}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{51}\right)\)
\(=\frac{1}{5}.\left(1-\frac{1}{51}\right)\)
\(=\frac{1}{5}.\left(\frac{51}{51}-\frac{1}{51}\right)\)
\(=\frac{1}{5}.\frac{50}{51}\)
\(=\frac{10}{51}\)
Chúc bạn học tốt !!!
\(A=\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+....+\frac{1}{496}-\frac{1}{501}\right):5\)
\(A=\left(1-\frac{1}{501}\right):5\)
\(A=\frac{500}{501}:5=\frac{100}{501}\)
Ta có : \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{496.501}\)
\(\Rightarrow\) \(A=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{496}-\frac{1}{501}\right) \)
\(\Rightarrow\) \(A=\frac{1}{5}\left(1-\frac{1}{501}\right)\)
\(\Rightarrow\) \(A=\frac{1}{5}.\frac{501-1}{501}=\frac{1}{5}.\frac{500}{501}\)
\(\Rightarrow\) \(A=\frac{1.500}{5.501}=\frac{20}{1.501}=\frac{20}{501}\)
Vậy \(A=\frac{20}{501}\)
