\(\frac{3}{4}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{99.100}\)
\(=\frac{3}{4}+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{3}{4}+1-\frac{1}{100}=\frac{87}{50}\)
Đề bài: Tính
\(\frac{3}{4}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{3}{4}+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{3}{4}+1-\frac{1}{100}\)
\(=\frac{3}{4}+\frac{99}{100}=\frac{75}{100}+\frac{99}{100}=\frac{174}{100}=\frac{87}{50}\)
\(\frac{3}{4}+\frac{1}{1.2}+...+\frac{1}{99.100}=\frac{3}{4}+\frac{2-1}{1.2}+....+\frac{100-99}{99.100}=\frac{3}{4}+1-\frac{1}{2}+...-\frac{1}{100}\)
\(=\frac{7}{4}-\frac{1}{100}=\frac{174}{100}\)