Question

\(\frac{2}{1.3}\)+  \(\frac{2}{3.5}\)+  \(\frac{2}{5.7}\)+...+ \(\frac{2}{n\left(n+2\right)}\)< \(\frac{2003}{2004}\)

Giúp mình tìm n \(\in\)\(ℤ\)với

 

Answer 1

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n\left(n+2\right)}\)

\(=\frac{1}{2}\left(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{n}-\frac{2}{n+2}\right)\)

\(=\frac{1}{2}\left(2-\frac{2}{n+2}\right)=\frac{1}{2}\cdot\frac{2n+2}{n+2}=\frac{n+1}{n+2}< \frac{2003}{2004}\)

\(\Rightarrow\hept{\begin{cases}n+1=2002\\n+2=2003\end{cases}}\Leftrightarrow n=2001\)