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a fost răspuns

Va rog frumos ajutati-ma repede.
Fie functia f:R->R si f(x)=2x-4
Calculati f(1)+f(2)+f(3)+...+f(2011)

Răspuns :

Ctinamaria31xxx
f(1)=2-4=-2
f(2)=4-4=0
f(3)=6-4=2
f(4)=8-4=4
f(5)=10-4=6

........
f(2009)=4018+2=4020
f(2010)=4020+2=4022
f(2011)=4022+2=4024

 f(1)+f(2)+f(3)+...+f(2011)
(-2+0)+2+4+6+8+.........+4020+4022+4024
(-2+0)+2(1+2+3+4+......2010+2011+2012)=

1+2+3+4+......2010+2011+2012=   Am aplicat Gauss
2012*(2012+1)/2=
2012*2013/2=
1006*2013=
2025078

(-2+0)+2*2025078
-2+4050156=
4050154

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