Question

CMR 2/1.3 + 2/3.5 + 2/5.7 +...+ 2/n(n+2) < 2003/2004 (. là dấu nha)

Answer 1

Sửa đề \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{n\left(n+2\right)}< \frac{2014}{2014}=1\)

Ta có :

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

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

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

\(=1-\frac{1}{n+2}+0\)

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

Vì \(1-\frac{1}{n+2}< 1\) nên\(\frac{2}{1.3}+\frac{1}{3.5}+...+\frac{1}{n\left(n+2\right)}< 1\left(đpcm\right)\)