\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n.\left(n+1\right)}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..+\dfrac{1}{n}-\dfrac{1}{n-1}\)

\(A=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}=\dfrac{1999}{2000}\Rightarrow n.2000=\left(n+1\right).1999\)

\(\Leftrightarrow1999n+n=1999n+1999\Rightarrow n=1999\)

We have :

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{1999}{2000}\)
= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{1999}{2000}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{1999}{2000}\)
=  \(1-\dfrac{1}{n+1}=\dfrac{1999}{2000}\)
= > \(\dfrac{1}{n+1}=1-\dfrac{1999}{2000}\)
<=> \(\dfrac{1}{n+1}=\dfrac{1}{2000}\)
=> n + 1 = 2000
n = 2000 - 1
n = 1999
Vậy n = 1999.

A=12+16+112+...+1n(n+1)=11.2+12.3+...+1n.(n+1)

A=1−12+12−13+..+1n−1n−1

A=1−1n+1=nn+1=19992000⇒n.2000=(n+1).1999

⇔1999n+n=1999n+1999⇒n=1999

We have :

12+16+112+...+1n.(n+1)=19992000

= 11.2+12.3+13.4+...+1n.(n+1)=19992000
= 1−12+12−13+13−14+...+1n−1n+1=19992000
=  1−1n+1=19992000
= > 1n+1=1−19992000
<=> 1n+1=12000
=> n + 1 = 2000
n = 2000 - 1
n = 1999
Vậy n = 1999.