已知f(x)=sin[π/3(x+1)]-(√3)cos[π/3(x+1)],f(1)+f(2)+f(3)…+f(2013)=?

问题描述:

已知f(x)=sin[π/3(x+1)]-(√3)cos[π/3(x+1)],f(1)+f(2)+f(3)…+f(2013)=?

f(x)=2(1/2sin[π/3(x+1)])]-√3/2cos[π/3(x+1)])=2sin(πx/3)则 f(6k) = 0,f(6k+1) = f(6k+2) = √3 f(6k+3) = 0,f(6k+4) = f(6k+5) = -√3连续6个相加为0,由 2006 = 334*6 + 2∴原式 = f(2005)+f(2006) = f(1)+...