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Fie functia f: R → R, f(x) = -x - 5. b)S=f(1)+f(2)+….+f(100)

Răspuns :

Răspuns:

f(1)=-5-1=-6

f(2)=-7

....

f(100)=-105

s=-(6+7+...105)

suma de termeni a unei progresii aritmetice cu ratia 1 si primul termen 6

nr.de termeni=(105-6)+1=100

s100=100(6+105)/2=50*111=5550

suma ceruta =-5550