có dạng này nhưng là số chẵn nhân chãn

a),b) Tính ra rồi chứng minh (dãy số viết theo quy luật)

Đây là toán lớp 1 hả??????????? Tớ  nghĩ là toán lớ 6 đấy!!!!!!
 

1-1/100 , ủng hộ mk nha

=>2S=2/1.3+2/3.5+....+2/99.100

ơ bạn nhầm đề bài à

\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}-2x=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-2x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)\(5\left(1-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(5.\frac{99}{100}-2x=\frac{1}{2}.\frac{98}{99}\)

\(\frac{99}{20}-2x=\frac{49}{99}\)

\(2x=\frac{99}{20}-\frac{49}{99}\)

\(2x=\frac{8821}{1980}\)

\(x=\frac{8821}{1980}:2\)

\(x=\frac{8821}{3960}\)

\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

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

\(2A=\frac{1}{1}-\frac{1}{101}\)

\(2A=\frac{100}{101}\)

\(2A=\frac{100}{101}\Rightarrow A=\frac{100}{101}:2\)

\(\Rightarrow A=\frac{50}{101}\)

xl câu hỏi có chút thay đổi mong các bn thông cảm

a)1/5.6+1/6.7+1/7.8+.......+1/99.100

= (1/5-1/6)+(1/6-1/7)+(1/7-1/8)+.....+(1/99-1/100)

= 1/5 - 1/100

= 19/100

b)2/1.3+2/3.5+2/5.7+.........+2/2013.2015

= (1/1-1/3)+(1/3-1/5)+(1/5-1/7)+.....+(1/2013+1/2015)

= 1/1 - 1/2015

= 2014/2015

\(a,\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{99.100}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{5}-\frac{1}{100}=\frac{20}{100}-\frac{1}{100}=\frac{19}{100}\)

\(b,\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\)

\(=\frac{1}{1}-\frac{1}{2015}=\frac{2015}{2015}-\frac{1}{2015}=\frac{2014}{2015}\)