Question

Fie f:R->,f(x)=2x+3 , calculati f(0)+f(1)+f(2)+...+f(30)

Answer 1

se inlocuie x cu 0,1,2,3...30
f(1)=5, f(2)=7, f(3) =9 deja se observa ca este o suma a termenilor unei progresii aritmetice cu r=2. s30= ( f(1)+f(30))*/2
f(30)=f(1)+29r=5+58=63
deci s(30)= (5+63)*30/2
s(30)=68*15=1020