问题 解答题
已知一列非零向
an
满足:
a1
=(x1y1),
an
=(xnyn)=
1
2
(xn-1-yn-1xn-1+yn-1)(n≥2)

(Ⅰ)证明:{|
an
|}
是等比数列;
(Ⅱ)求向量
a
n-1
a
n
的夹角(n≥2)

(Ⅲ)设
a
1
=(1,2),把
a1
a2
,…,
an
,…中所有与
a1
共线的向量按原来的顺序排成
一列,记为
b1
b2
,…,
.
bn
,…,令
OB
n
=
b1
+
b2
+…+
bn
,0
为坐标原点,求点列{Bn}的极限点B的坐标.
(注:若点Bn坐标为(tnsn),且
lim
n→∞
tn=t,
lim
n→∞
sn=s,则称点B(t,s)为点列{Bn}
的极限点.)
答案

(I)|

an
|=
1
2
(xn-1-yn-1)2+(xn-1+yn-1)2

=

2
2
x2n-1
+
y2n-1
=
2
2
|
a
n-1
|,(n≥2),首项|
a1
|=
x21
+
y21
≠0,
|
an
|
|
a
n-1
|
=
2
2
为常数,∴{|
an
|}
是等比数列.

(II)

a
n-1
a
n
=(xn-1yn-1)•
1
2
(xn-1-yn-1xn-1+yn-1)=
1
2
(
x2n-1
+
y2n-1
)=
1
2
|
a
n-1
|2
cos<
a
n-1
a
n
>=
a
n-1
a
n
|
a
n-1
||
a
n
|
=
1
2
|
a
n-1
|2
|
a
n-1
|•
2
2
|
a
n-1
|
=
2
2
,∴
a
n-1
a
n
的夹角为
π
4

(III)

a1
=(x1y1),
a2
=
1
2
(x1-y1x1+y1),

a3
=
1
4
(-2y1,2x1)=
1
2
(-y1x1),
a4
=
1
4
(-y1-x1,-y1+x1),
a5
=
1
8
(-2x1,-2y1)=-
1
4
(x1y1)
,∴
a1
a5
a9
一般地,
b1
=
a1
b2
=
a5
,,
bn
=
a
4n-3

用数学归纳法易证

b
n=
a
4n-3
成立∴
b
n
=(-
1
4
)n-1(x1y1)

OBn
=(tnsn)则tn=[1+(-
1
4
)+(-
1
4
)
2
+…+(-
1
4
)
n-1
]x1=
1-(-
1
4
)
n
1-(-
1
4
)
=
4
5
[1-(-
1
4
)n],
lim
n→∞
tn=
4
5

sn=[1+(-

1
4
)+(-
1
4
)
2
+…+(-
1
4
)
n-1
]y1=
1-(-
1
4
)
n
1-(-
1
4
)
•2=
8
5
[1-(-
1
4
)n],
lim
n→∞
sn=
8
5

∴极限点B的坐标为(

4
5
8
5
).

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