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2014-07-14T13:19:07+03:00
F(0)=2·0+3=3
f(1)=2·1+3=5
f(2)=2·2+3=7
f(3)=2·3+3=9
f(4)=2·4+3=11
f(5)=2·5+3=13
_____________________

f(0)+f(1)+f(2)+f(3)+f(4)+f(5)=
=3+5+7+9+11+13
=48

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2014-07-14T13:26:38+03:00
F(0) =2x0+3=3
f(1)= 2x1+3=5
f(2)= 2x2+3=7
f(3)= 2x3+3=9
f(4)=2x4+3=11
f(5)=2x5+3=13
_________________+
3+5+7+9+11+13=48



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