答：在0~2π周期内：sin(π/4)+sin(2π/4)+sin(3π/4)+sin(4π/4)+.+sin(8π/4)=0100/8=12余4f(x)=sin(πx/4)f(0)+f(1)+f(2)+.+f(100)=sin0+sin(π/4)+sin(2π/4)+sin(3π/4)+.+sin(100π/4)=0+0+.+sin(97π/4)+sin(...答案是1+跟号2请重新做好吗=sin(π/4)+sin(2π/4)+sin(3π/4)+sin(4π/4)=√2/2+1+√2/2+0……………………………（这里弄错了sin(2π/4)=1，原来弄成了2，sorry）=1+√2