[tex]\\ $Alta varianta, cea clasica: \\ \\ $ f:\mathbb_{R} \rightarrow \mathbb_{R}, $ $\quad f(x) = 2x+3. \\ \\ f(0)+f(1)+...+f(5) =\sum\limits_{x=0}^{5}f(x) =\sum\limits_{x=0}^{5}(2x+3) = \sum\limits_{x=0}^{5}(2x)+\sum\limits_{x=0}^{5}3 = \\ \\ =2\cdot \sum\limits_{x=0}^{5}x + 3\cdot \sum\limits_{x=0}^51 = \\ \\ = 2\cdot (0+1+2+...+5)+ 3\cdot(1+\underset{de \quad 6 \quad ori}{...}+1) = \\ \\ =2\cdot \dfrac{5\cdot6}2}+3\cdot 6 =5\cdot 6+3\cdot 6 = 6\cdot(5+3) = 6\cdot 8 = 48[/tex]
