Question

Fie funcția f:R➡R,f(x)=2x-1. Sa se calculeze: f(1)+f(2)+f(3)+....+f(2018)

Answer 1

Dupa al treilea rand era :2(0+1+2+3+...+2018)-1•2018=
=2•2018•2019:2-2018=
=4074342-2018=
=4072324
Sper ca te-am ajutat !

Answer 2

f(1)=2×1-1
f(2)=2×2-1
f(3)=2×3-1...f(2018)=2×2018-1
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S=2(1+2+3+........2018)-1-1-1-.............-1 (de 2018 ori)
S=2·2018·2019/2- 2018
S=2018·2019-2018
S=2018·(2019-1) ⇒S=2018·2018 ⇒S=2018²