Question

Tìm số nguyên n biết :

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\)

Answer 1

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2001}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1999}{2001}\)

\(2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{1999}{2001}\)

\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)

\(\frac{1}{n+1}=\frac{1}{2001}\)

=>n+1=2001

=>n=2000