Question

tìm n thuộc N:1/3+1/6+1/10+...+2/n.(n+1)=2003/2004

Answer 1

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{n\left(n+1\right)}=\frac{2003}{4008}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{2003}{4008}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)

\(=\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)

\(\frac{1}{n+1}=\frac{1}{4008}\)

\(\Rightarrow\)n+1=4008

n=4007

Vậy n=4007

Answer 2

 TA CÓ :\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{n\left(n+1\right)}\)\(=\frac{2003}{2004}\)

      \(Nhân\)\(cả\)\(hai\)\(vế\)\(với\)\(\frac{1}{2}\),  TA ĐƯỢC :

    \(\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{n.\left(n+1\right)}\right)\)\(=\frac{1}{2}.\frac{2003}{2004}\)
=>\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{n.\left(n+1\right)}\)\(=\frac{2003}{4008}\)

=>\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{n.\left(n+1\right)}\)\(=\frac{2003}{4008}\)

=>\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{n}-\frac{1}{n+1}\)\(=\frac{2003}{4008}\)

=>\(\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)

=>\(\frac{1}{n+1}=\frac{1}{4008}\)

=> \(n+1=4008\)

=> \(n=4007\)( Thỏa mãn điều kiện : \(n\in N\))

Vậy n=4007