Question

Se considera functia f:R-> R ,f(x)=2x+3. Sa se calculeze f(0) +f(1)+......+f(5).

Answer 1

f:R→R
f(x)=2x+3

f(0)=2*0+3=3
f(1)=2*1+3=2+3=5
f(2)=2*2+3=4+3=7
f(3)=2*3+3=6+3=9
f(4)=2*4+3=8+3=11
f(5)=2*5+3=10+3=13

 f(0)+f(1)+f(2)+f(3)+f(4)+f(5)=3+5+7+9+11+13=48

Answer 2

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