已知一列非零向量
(Ⅰ)证明:{|
(Ⅱ)求向量
(Ⅲ)设
(注:若点Bn坐标为(tn,sn),且
|
(I)|
|=an 1 2 (xn-1-yn-1)2+(xn-1+yn-1)2
=
•2 2
=
+x 2n-1 y 2n-1
|2 2
n-1|,(n≥2),首项|a
|=a1
≠0,
+x 21 y 21
=|
|an |
n-1|a
为常数,∴{|2 2
|}是等比数列.an
(II)
n-1•a
n=(xn-1,yn-1)•a
(xn-1-yn-1,xn-1+yn-1)=1 2
(1 2
+x 2n-1
)=y 2n-1
|1 2
n-1|2,cos<a
n-1,a
n>=a
=
n-1•a
na |
n-1||a
n|a
=
|1 2
n-1|2a |
n-1|•a
|2 2
n-1|a
,∴2 2
n-1与a
n的夹角为a
.π 4
(III)
=(x1,y1),a1
=a2
(x1-y1,x1+y1),1 2
=a3
(-2y1,2x1)=1 4
(-y1,x1),1 2
=a4
(-y1-x1,-y1+x1),1 4
=a5
(-2x1,-2y1)=-1 8
(x1,y1),∴1 4
∥a1
∥a5
∥一般地,a9
=b1
,a1
=b2
,,a5
=bn
4n-3,a
用数学归纳法易证
n=b
4n-3成立∴a
n=(-b
)n-1(x1,y1).1 4
设
=(tn,sn)则tn=[1+(-OBn
)+(-1 4
)2+…+(-1 4
)n-1]x1=1 4
=1-(-
)n1 4 1-(-
)1 4
[1-(-4 5
)n],1 4
tn=lim n→∞
;4 5
sn=[1+(-
)+(-1 4
)2+…+(-1 4
)n-1]y1=1 4
•2=1-(-
)n1 4 1-(-
)1 4
[1-(-8 5
)n],1 4
sn=lim n→∞
,8 5
∴极限点B的坐标为(
,4 5
).8 5