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Se considera functia f:R->R f(x)=4x-5 .Calculați f(1)+f(2)+f(3)+...+f(10)

Răspuns :

f:R->R
f(x)=4x-5
f(1)= 4*1-5=-1
f(2)=4*2-5=3
f(3)=12-5=7
f(10)=35
formeaza o progresie aritmetica cu a1=-1 si r=4 n=10
S=(a1+an)*n/2
an=a1+(n-1)*r= -1+9*4=35
S=(-1+35)*10/ 2=170