[tex] \frac{1}{1\cdot4}+ \frac{1}{4\cdot7}+ \frac{1}{7\cdot10}+...+ \frac{1}{100\cdot103}= \\ \\ \frac{1}{3}( \frac{1}{1}- \frac{1}{4})+ \frac{1}{3}( \frac{1}{4}- \frac{1}{7})+ \frac{1}{3}( \frac{1}{7}- \frac{1}{10})+...+ \frac{1}{3}( \frac{1}{100}- \frac{1}{103})= \\ \\ \frac{1}{3}( \frac{1}{1}- \frac{1}{4}+ \frac{1}{4}- \frac{1}{7}+ \frac{1}{7}- \frac{1}{10}+...+ \frac{1}{100}- \frac{1}{103})= \\ \\ \frac{1}{3}( \frac{1}{1}- \frac{1}{103})= \\ \\ \frac{1}{3}\cdot \frac{102}{103} [/tex]
-deci aici am folosit faptul ca n(n+k)=k[n+(n+k)] o modificare a sumei gauss deci am descompus fractiile apoi le-am simplificat si obtinem ca [tex] \frac{1}{3}\cdot \frac{102}{103}= \frac{34}{103}[/tex] acesta este rezultatul.
