\(B=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{1}{55}\)

\(B=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{110}\)

\(B=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{10.11}\)

\(B=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(B=2.\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)

làm cả 3 nhé 

bạn do not ask why làm sai ở bước 4 rồi 

6F=1.3(5-1)+3.5(7-1)+5.7(9-3)+...99.101(103-97)

6F=1.3+1.3.5-1.3.5+3.5.7-3.5.7+.....-97.99.101+99.101.103

6F=3+99.101.103

6F=3+1029897

6F=1029900

F =1029900:6

F=171650

a, \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

\(=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{101-99}{99.101}\)

\(=\dfrac{3}{1.3}-\dfrac{1}{1.3}+\dfrac{5}{3.5}-\dfrac{3}{3.5}+...+\dfrac{101}{99.101}-\dfrac{99}{99.101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

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

b, \(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=\dfrac{5}{2}.\left(1-\dfrac{1}{101}\right)=\dfrac{5}{2}.\dfrac{100}{101}=\dfrac{500}{202}=\dfrac{250}{101}\)

a)

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+..............+\dfrac{2}{99.101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...............+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

b)

\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+......................+\dfrac{5}{99.101}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...............+\dfrac{2}{99.101}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+............+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)

\(=\dfrac{5}{2}.\dfrac{100}{101}=\dfrac{250}{101}\)

\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}.\)\(B=\dfrac{5}{1.3}+\dfrac{5}{3.5}+...+\dfrac{5}{99.100}=\dfrac{5}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)=\dfrac{5}{2}.A=\dfrac{5}{2}.\dfrac{100}{101}=\dfrac{250}{101}.\)