Question

定义在r上的函数f(X)满足f(x+6)=f(X).当x属于[-3,1]时，f(X)=-(x+2)^2,当x属于[-1,3]时f(X)=x.

则f(1)+f(2)+f(3)+f(4)+........+f(2012)=

Answer 1

T=6
2012/6=335……2
f(1)+f(2)+f(3)+f(4)+........+f(2012)=335(f1+f2+f3+f4+f5+f6)+f1+f2
f1=1
f2=2
f3=3
f4=f-2=0
f5=f-1=-1
f6=f0=0
f1+f2+f3+f4+f5+f6=5

f(1)+f(2)+f(3)+f(4)+........+f(2012)=5*335+1+2=1678

Answer 2

T=6
2012/6=335……2
f(1)+f(2)+f(3)+f(4)+........+f(2012)=335(f1+f2+f3+f4+f5+f6)+f1+f2
f1=1
f2=2
f3=-1
f4=f-2=0
f5=f-1=-1
f6=f0=0
f1+f2+f3+f4+f5+f6=1
f(1)+f(2)+f(3)+f(4)+........+f(2012)=1*335+1+2=338

Answer 3

f(3)是等于–1.答案是B.338