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25+3t>015分25综上解得:1>17分1219.(17分)解(1)由题意可得:x=a·a=-2,y=b·a=(2,0)·(2,-1)=4a=(-2,4))[a=2√54分(2)a之间的等量关系为axa(k N).6分a=√x²+y²,ax|=√x+²+yx+²axa√x+y√x+y√xx+xy+x+yy分又bk+1=(yk+1-xk+1)√xk+2=ak+1·ak=xk+1xk+yk+1yk.8分yk+2=bk+1·ak=xkyk+1-xk+1ykak+2=(xxxk+1+ykyk+1xkyk+1-xk+1yk)ak+2=√(xxk+1+ykyk+1)²+(xxyk+1-xk+1yx)²=√xx+1+xy+xx+y+yy+故ax+2=axa(k ∈ N").10分(3)证明:由(2)及=a=1,可得=1,依次类推a=1(k∈N),1分可设ak=(cosθk,sinθk),则ak+1=(cos θk+1Sin0k+1),bk+=(sinθk+1-cosθk+1)12分依题意得:xk+2=ak+1·ak=cosθk+1cos0k+sin0k+1sinθk=cos(0k+1-0k)yk+2=bk+1·ak = sin0k+1cos θk-cos θk+1 sinθk = sin(0k+1-0).ak+2=(cos(θk+1-0k),sin(0k+1-0k))14分同理得:αak+3 =(cos[(0k+1 -0k)-0k+1]sin[(0k+1-0)-0k+1])=(cos(-0),sin(-0))依次类推可得:ak+4=(cos(-0k+1),sin(-0k+1))ak+s=(cos(0k-0k+1),sin(0k-0k+1))
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