Question

CMR\(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}< 1\)

Answer 1

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

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\Rightarrow\frac{99}{100}< 1\)

~Học tốt~

Answer 2

Ở đây đề bài phải là :

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

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

=\(1-\frac{1}{100}< 1\)

Vậy \(A< 1\)

Answer 3

\(\frac{1}{2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{99\cdot100}\)

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

\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)