Ta có :
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.................+\dfrac{2}{n\left(n+1\right)}=\dfrac{2003}{2004}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+............+\dfrac{2}{n\left(n+1\right)}=\dfrac{2003}{2004}\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+............+\dfrac{2}{2\left(n+1\right)}=\dfrac{2003}{2004}\)
\(2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{2003}{2004}\)
\(2\left(\dfrac{1}{2}-\dfrac{1}{n+1}\right)=\dfrac{2003}{2004}\)
\(\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{2003}{4008}\)
\(\dfrac{1}{n+1}=\dfrac{1}{4008}\)
\(\Rightarrow n+1=4008\)
\(\Rightarrow n=4007\) (Thỏa mãn \(n\in N\))
Vậy \(n=4007\) là giá trị cần tìm
~~Chúc bn học tốt~~
