Question

求和：1+cosx+cos2x+…+cosnx 希望大神指点迷津～谢

Answer 1

对于Sn=cosx+cos2x+cos3x+……+cosnx，有：
2sin(x/2)[cosx+cos2x+cos3x+……+cosnx ]
=2sin(x/2)cosx+2sin(x/2)cos2x+2sin(x/2)cos3x+……+2sin(x/2)cosnx
=sin(3x/2)-sin(x/2)+sin(5x/2)-sin(3x/2)+sin(7x/2)-sin(5x/2)+……+sin(x/2+nx)-sin(nx-x/2)
=sin(x/2+nx)-sin(x/2)
所以，
cosx+cos2x+cos3x+……+cosnx
=[sin(x/2+nx)-sin(x/2)]/[2sin(x/2)]
=sin(x/2+nx)/[2sin(x/2)]-1/2

故：1+cosx+cos2x+…+cosnx = sin(x/2+nx)/[2sin(x/2)] + 1/2