设k∈Nsin[(4k+1)π/2]+cos[(4k+1)π/2]=1+0=1sin[(4k+2)π/2]+cos[(4k+2)π/2]=0-1=-1sin[(4k+3)π/2]+cos[(4k+3)π/2]=-1+0=-1sin[(4k+4)π/2]+cos[(4k+5)π/2]=0+1=1因此sin(nπ/2)+cos(nπ/2),n从1开始,按1,-1,-...