已知f(n)=1-sin(nπ/6 ),求f(1)+f(2)+·····+f(2011)谢谢

f(1)=1/2
f(2)=1-根号3/2
f(3)=0
f(4)=1-根号3/2
f(5)=1/2
f(6)=1
f(7)=3/2
f(8)=1+根号3/2
f(9)=2
f(10)=1+根号3/2
f(11)=3/2
f(12)=1
f(n)周期为12.
f(n+12)=f(n)
f(1)+f(2)+...+f(12)=12
f(1)+f(2)+·····+f(2011)=167*12+f(2005)+f(2006)+f(2007)+...+f(2011)
=2004+f(1)+f(2)+...+f(7)=2004+5.5-根号3=2009.5-根号3