问题 解答题
已知一非零的向量列
an
满足:
a1
=(1,1),
an
=(xnyn)=
1
2
(xn-1-yn-1xn-1+yn-1)(n≥2)

(1)计算|
a1
|,|
a2
|,|
a3
|;证明:数列{|
an
|}是等比数列;
(2)设θn(n≥2)是
an-1
an
的夹角的弧度数,bn=
π
4n(n-1)θn
Sn=b2+b3+…+bn,求S2013
答案

(1)∵非零向量列{

an
}满足:
a1
=(1,1),
an 
=(xn,yn)=
1
2
(xn-1-yn-1,xn-1+yn-1) (n≥2),

a1
=(1,1),
a2
=
1
2
(0,2)=(0,1),
a3
=
1
2
(-1,1)=(-
1
2
1
2
),

∴|

a1
|=
12+12
=
2
,|
a2
|=
02+12
=1,|
a3
|=
(-
1
2
)2+(
1
2
)2
=
2
2

∵|

an
|=
xn2+yn2

∴|

an+1
|=
xn+12+yn+12
=
(
xn-yn
2
)2+(
xn+yn
2
)2
=
2
2
xn2+yn2

|
an+1
|
|
an
|
=
2
2
(常数),

∴{|

an
|}是首项|
a1
|=
2
,公比q=
2
2
的等比数列.

(2)∵

an-1
an
=(xn-1,yn-1)•
1
2
(xn-1-yn-1,xn-1+yn-1

=

1
2
xn-12+yn-12)=
1
2
|
an-1
|2

∴cosθn=

an-1
an
|
an-1
|•|
an
|
=
1
2
|
an-1
|
|
an
|
=
2
2

∴θn=

π
4
,n≥2.

∴bn=

π
4n(n-1)θn
=
π
4n(n-1)•
π
4
=
1
n(n-1)
=
1
n-1
-
1
n

∴Sn=b2+b3+…+bn

=(1-

1
2
)+(
1
2
-
1
3
)+…+(
1
n-1
-
1
n

=1-

1
n

∴S2013=1-

1
2013
=
2012
2013

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