Question

\(CMR\)\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}< 1\)

Answer 1

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(=1-\frac{1}{2020}< 1\)

\(\Rightarrow A< 1\left(đpcm\right)\)

Answer 2

\(A=1-\frac{1}{2}+\frac{1}{2}-...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=1-\frac{1}{2020}\)

\(=>ĐPCM\)

Answer 3

Ta chứng minh được \(\frac{1}{m\left(m+1\right)}=\frac{m+1-m}{m\left(m+1\right)}=\frac{m+1}{m\left(m+1\right)}-\frac{m}{m\left(m+1\right)}=\frac{1}{m}-\frac{1}{m+1}\)

Ta có \(\frac{1}{1.2}=1-\frac{1}{2}\)

      \(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

     .....

        \(\frac{1}{2019.2020}=\frac{1}{2019}-\frac{1}{2020}\)

=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)

=>\(A=1-\frac{1}{2020}< 1\)

Vậy \(A< 1\)

Answer 4

Ta có:

A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2019.2020}\)

A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2019}-\frac{1}{2020}\)

A = \(1-\frac{1}{2020}\) 
A = \(\frac{2019}{2020}\) < 1