\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+.....+\dfrac{5}{100.103}\)
\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(B=\dfrac{5}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(B=\dfrac{5}{3}.\left(1-\dfrac{1}{103}\right)\)
\(B=\dfrac{5}{3}.\dfrac{102}{103}\)
\(B=1\dfrac{67}{103}\)

Ta có: \(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)

\(=\dfrac{5}{3}\left(1-\dfrac{1}{4}\right)+\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}\right)+...+\dfrac{5}{3}\left(\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{3}.\dfrac{102}{103}=\dfrac{170}{103}\)

Vậy \(B=\dfrac{170}{103}\).

Ta có : B = \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)

\(\Rightarrow B=\dfrac{5}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)

\(\Rightarrow B=\dfrac{5}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(\Rightarrow B=\dfrac{5}{3}.\left(1-\dfrac{1}{103}\right)\)

\(\Rightarrow B=\dfrac{5}{3}.\dfrac{102}{103}\)

\(\Rightarrow B=\dfrac{170}{103}\)

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\(\dfrac{3}{2}\)B= \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)

68/103

\(\dfrac{1}{3}\)x(\(\dfrac{3}{1+4}\)+\(\dfrac{3}{4+7}\)+........+\(\dfrac{3}{101+103}\))

\(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+.........+\(\dfrac{ }{ }\)\(\dfrac{1}{101}\)-\(\dfrac{1}{103}\))

\(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{103}\))

\(\dfrac{1}{3}\)x\(\dfrac{102}{103}\)

\(\dfrac{34}{103}\)

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=\dfrac{1}{1}-\dfrac{1}{103}\)

\(=\dfrac{102}{103}\)

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)=\dfrac{1}{3}.\dfrac{102}{103}=\dfrac{102}{309}=\dfrac{34}{103}\)

A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)

=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)

=2.(1-1/101)

=2.(101/101-1/101)

=2.100/101

200/101

B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)

=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)

=2.(1/1+1/101)

=2.(101/101+1/101)

=2.102/101

=204/101

C=1/2+1/3+1/3+1/4+....+1/99+1/100

=1/2+1/100

=50/100+1/100

=51/100

a) Ta có: \(A=\dfrac{4}{1\cdot4}+\dfrac{4}{4\cdot7}+\dfrac{4}{7\cdot10}+...+\dfrac{4}{31\cdot34}\)

\(=\dfrac{4}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{31\cdot34}\right)\)

\(=\dfrac{4}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)\)

\(=\dfrac{4}{3}\left(1-\dfrac{1}{34}\right)\)

\(=\dfrac{4}{3}\cdot\dfrac{33}{34}=\dfrac{22}{17}\)

\(B=\dfrac{4}{1.4}+\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{100.103}+\dfrac{4}{103.106}\)

\(B=4\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}+\dfrac{1}{103.106}\right)\)

\(B=\dfrac{4}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}+\dfrac{3}{103.106}\right)\)

\(B=\dfrac{4}{3}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}+\dfrac{1}{103}-\dfrac{1}{106}\right)\)

\(B=\dfrac{4}{3}\left(\dfrac{1}{3}-\dfrac{1}{106}\right)\)

\(B=\dfrac{4}{3}.\dfrac{103}{318}\)

\(B=\dfrac{412}{954}\)

Câu 1 :

1/n - 1/n + a = a + n/a ( a + n ) = a + n - a/a ( n + a ) = n/a ( a + n )

Câu 2 :

A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +.......+ 1/99 - 1/100

= 1/1 - 1/100 = 99/100

a, \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{n+a-n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)

Vậy \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{a}{n\left(n+a\right)}\)

b,

\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)

\(3B=\dfrac{5.3}{1.4}+\dfrac{5.3}{4.7}+...+\dfrac{5.3}{100.103}\)

\(3B=5\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)

\(3B=5\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(3B=5\left(1-\dfrac{1}{103}\right)=5\cdot\dfrac{102}{103}=\dfrac{510}{103}\)

\(B=\dfrac{510}{103}:3=\dfrac{170}{103}\)

\(C=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)

\(C=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)

\(2C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\)

\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\)

\(2C=\dfrac{1}{3}-\dfrac{1}{51}=\dfrac{16}{51}\)

\(C=\dfrac{16}{51}:2=\dfrac{8}{51}\)

=\(2.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+.....+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

=\(2.\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\)

= \(2.\dfrac{99}{100}\)

=\(\dfrac{99}{50}\)