Question

Fie functia f : R ⇒ R, f(x) = 2x - 8. Calulati:
a) f(1) + f(2) + f(3) + f(4) + f(5) + ... + f(2014)
b) f(1) · f(2) · f(3) · f(4) · f(5) · ... · f(2014) ·

Answer 1

a) f(1) + f(2) + f(3) + f(4) + f(5) + ... + f(2014) =-6 -4-2+0+2 +4+6+ ... + 4048=

= -12+2(1+2+3+ ... + 2012) = -12+ 2012· 2013

b) f(1) · f(2) · f(3) · f(4) · f(5) · ... · f(2014)  = 0,  pentru că f(4) =0

Answer 2

a)f(1)= - 6, f(2)= - 4, f(3)= - 2 , f(4)=0 => f(2)-f(1)= 2, f(3)-f(2)=2 => este progresie aritmetica
S=[(a1+an)*n]/2=>S=[(-6+4020)]*2014]/2=4014*1007=4042098
f(2014)=4020
n=2014
b)Cum am vazut in a) f(4)=0 =>

f(1) · f(2) · f(3) · f(4) · f(5) · ... · f(2014)=0

coroana