Question

[Alexandra]

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

GIÚP MÌNH NHA MÌNH SẮP NỘP RÙI

Answer 1

[Nguyễn Đình Dũng]

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

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

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

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

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

Answer 2

[Lightning Farron]

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

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

Ta thấy:

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

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

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

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

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

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