Question

已知函数f(x)=sinπx/3,则f(1)+f(2)+---+f(2014)=

Answer 1

f(x)=sinπx/3,
最小正周期T=2π/(π/3)=6
f(1)=sinπ/3=√3/2
f(2)=sin2π/3=√3/2
f(3)=sinπ=0

f(4)=sin4π/3=-√3/2
f(5)=sin5π/3=-√3/2
f(6)=sin2π=0
1个周期中f(1)+f(2)+f(3)+f(4)+f(5)+f(6)=0
又2014=6×335+2

f(1)+f(2)+--+(6)+f(7)+---+f(2012)+f(2013+f(2014)
=f(1)+f(2)
=√3

Answer 2

解：
易知
f(1)+f(4)=sinπ/3+sin4π/3=0
f(2)+f(5)=sin2π/3+sin5π/3=0
f(3)+f(6)=sin3π/3+sin6π/3=0

而f(1)=f(7)=...=f(1+6×334)=sinπ/3
f(2)=f(8)=...f(2+6×334)=sin2π/3
...
f(6)=f(12)=...=f(6+6×334)=sin6π/3

故
f(1)+f(2)+---+f(2014)
=[f(1)+f(2)+f(3)+f(4)+f(5)+f(6)]×335+f(2011)+f(2012)+f(2013)+f(2014)
=0×335+f(2012)+f(2013)
=0+sin2π/3+sinπ
=√3/2