Đặt \(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{99.100}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=2.\left(1-\frac{1}{100}\right)\)
\(A=\frac{2.99}{100}\)
\(A=\frac{99}{50}=1\frac{49}{50}\)
Đúng 0
Bình luận (0)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)=2.\frac{99}{100}\)
\(=\frac{99}{50}\)
Đúng 0
Bình luận (0)