=1+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{2}\) -\(\frac{1}{3}\) -\(\frac{1}{4}\)+\(\frac{1}{3}\) - \(\frac{1}{4}\)-\(\frac{1}{5}\)+.....+\(\frac{1}{99}\)-\(\frac{1}{100}\)-\(\frac{1}{101}\)

=1+\(\frac{1}{101}\)

=\(\frac{102}{101}\)

1/1.2.3 = 1/2 .[1/1.2 - 1 / 2.3]

1/2.3.4 = 1/2[ 1/2- 1/3 ] 

...................

1/99.100.101 = 1/2[ 1/99. 100 - 1/100.101]

=> A= 1/2 [ 1/1.2- 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/ 4.5 +.........+ 1/99 .100 - 1/100. 101]

A = 1/2 . [1/1.2 -1/100 .101]

A= 1/2 . 5049 /10100 = 5049 / 20200.

Mình nghĩ là vậy đó.

a=100/101

A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+..+\frac{1}{99.100.101}\)

A = \(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{101-99}{99.100.101}\right)\)

A = \(\frac{1}{2}.\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+...+\frac{101}{99.100.101}-\frac{99}{99.100.101}\right)\)

A = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

A = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100.101}\right)\)

A = \(\frac{1}{2}.\frac{5049}{10100}\)

A = \(\frac{5049}{20200}\)

\(A=\frac{5049}{20200}\)

A=\(\frac{5049}{20200}\)

A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)

=> A = \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)

= \(\frac{1}{2}.\frac{5049}{10100}\)

= \(\frac{5049}{20200}\)

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\)

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{99.100.101}\)

Ta thấy:

\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3};\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4};...;\frac{2}{99.100.101}=\frac{1}{99.100}-\frac{1}{100.101}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{100.101}\)

\(\Rightarrow2A=\frac{1}{2}-\frac{1}{10100}\)

\(\Rightarrow2A=\frac{5050}{10100}-\frac{1}{10100}\)

\(\Rightarrow2A=\frac{5049}{10100}\Rightarrow A=\frac{5049}{10100}:2=\frac{5049}{20200}\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)

\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)

\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)

\(C=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+....+\frac{101-99}{99.100.101}\)

\(C=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}+\frac{2}{100.101}\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)

\(C=\frac{1}{2}\cdot\frac{5049}{10100}=\frac{5049}{20200}\)

Bài này hơi dài nên bạn tham khảo tại đây nha :

Câu hỏi của Kim Sura xXx pÉ heO - Toán lớp 6 - Học toán với OnlineMath

Phân tích:

\(\frac{1}{1.2.3}=1-\left(\frac{1}{2}+\frac{1}{3}\right)\)

\(\frac{1}{2.3.4}=\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)\)

Phân tích tương tự với hai số kia

Ta được: C = \(1-\left(\frac{1}{2}+\frac{1}{3}\right)+\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)+\frac{1}{3}-\left(\frac{1}{4}+\frac{1}{5}\right)+...+\frac{1}{99}-\left(\frac{1}{100}+\frac{1}{101}\right)\)

= \(1-\frac{1}{101}\)

= \(\frac{100}{101}\)