O inegalite de cls a sasea
Posted: Sat Nov 08, 2008 9:27 am
Demonstrati ca : \( \frac{25}{2\cdot7}+\frac{25}{7\cdot12}+\frac{25}{12\cdot17}+...+\frac{25}{47\cdot52}<\frac{5}{2} \)
Pentru inceput: \( \frac{1}{k(k+5)}=\frac{1}{5}(\frac{1}{k}-\frac{1}{k+5}) \), deci \( \frac{1}{2 \cdot 7}+\frac{1}{7 \cdot 12}+...+\frac{1}{47 \cdot 52}=\frac{1}{5}(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{47}-\frac{1}{52})=\frac{1}{5}(\frac{1}{2}-\frac{1}{52})=\frac{1}{5} \cdot \frac{25}{52}=\frac{5}{52} \).
Rezulta ca \( LHS=25(\frac{1}{2 \cdot 7}+\frac{1}{7 \cdot 12}+...+\frac{1}{47 \cdot 52}=25 \cdot \frac{5}{52}=\frac{125}{52}<\frac{2}{5}(\Longleftrightarrow 250<260). \)
Observatie. \( \text{LHS}=\text{Left Hand Side}=\text{Membrul Stang} \)