Question

f:R > R, f(x)= - 3x+2.Sa se calculeze:
f(1)+f(2)+...+f(9).

Answer 1

$f:\mathbb{R} \to \mathbb{R},~f(x)-3x+2 \\ \\ \text{S\u{a} se calculeze: }f(1)+f(2)+...+f(9).\\ \\ \\ f(1)+f(2)+...+f(9) = \sum\limits_{x=1}^{9} f(x) = \sum\limits_{x=1}^{9}(-3x+2) =\\ \\ = \sum\limits_{x=1}^9(-3x)+\sum\limits_{x=1}^{9} 2 = -3\cdot \sum\limits_{x=1}^9 x+ 2\cdot 9 = -3\cdot \dfrac{9\cdot (9+1)}{2} + 18 = \\ \\ =-3\cdot \dfrac{9\cdot 10}{2}+18 = -3\cdot 9\cdot 5+18 =\boxed{-117}$

Answer 2

Cand am transformat acea suma am folosit formula progresiei aritmetice.