Question

已知奇函数fx(x∈R),满足f(x＋4)=fx＋f2,且f1=2,则f1＋f2＋f3＋…＋f2

11是多少

Answer 1

f(x)是奇函数，f(0)=-f(0)
∴f(0)=0
f(x+4)=f(x)+f(2)
f(-2+4)=f(2)=f(-2)+f(2)=-f(2)+f(2)
∴f(2)=0
f(x)=f(x+4)
f(x)是最小正周期为4的周期函数
f(3)=f(-1+4)=f(-1)=-f(1)=-2
f(4)=f(0)=0
f(5)=f(1)=2
...
∴f1＋f2＋f3＋…＋f211
=(2+1-2+0)×52+2+1-2
=53