\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\) <=>\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+.....\frac{1}{n}-\frac{1}{n-1}\right)=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{1999}{2001}\)
<=>\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)
<=>\(\frac{1}{n+1}=\frac{1}{2001}\)
<=>n+1 =2001
<=>n = 2000
ta có:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{2001}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}\right)=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{1}{2.3}+\frac{1}{2.6}+\frac{1}{2.10}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)
\(\frac{1}{n+1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\frac{1}{n+1}=\frac{1}{2001}\)
=>\(n+1=2001\)
=>\(n=2000\)
Ta có :
\(A.\frac{1}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+..............+\frac{1}{n.\left(n+1\right)}=\frac{1999}{2001}\)
=> \(A.\frac{1}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..............+\frac{1}{n.\left(n+1\right)}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..............+\frac{1}{n}-\frac{1}{n+1}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\Rightarrow\frac{1}{n+1}=\frac{1}{2}-\frac{1999}{4002}=\frac{1}{2001}\)
\(\Rightarrow n+1=2001\Rightarrow n=2001-1=2000\)
Vậy n = 2000
