【答案】
分析:(I)由題意知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182351367169176/SYS201310241823513671691026_DA/0.png)
,a
4=(1+cos
2π)a
2+4sin
2π=2a
2=4,一般地,當(dāng)n=2k-1(k∈N
*)時(shí),a
2k+1-a
2k-1=4.因此a
2k-1=4(k-1).當(dāng)n=2k(k∈N
*)時(shí),a
2k=2
k.由此可知數(shù)列{a
n}的通項(xiàng)公式為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182351367169176/SYS201310241823513671691026_DA/1.png)
.
(II)由題設(shè)知,S
k=a
1+a
3++a
2k-1=0+4++4(k-1)=2k(k-1),T
k=a
2+a
4++a
2k=2+2
2+2
k=2
k+1-2,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182351367169176/SYS201310241823513671691026_DA/2.png)
由此可知當(dāng)k≥6時(shí),W
k+1<W
k.滿足W
k>1的所有k的值為3,4,5.
解答:解:(I)因?yàn)閍
1=0,a
2=2,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182351367169176/SYS201310241823513671691026_DA/3.png)
,a
4=(1+cos
2π)a
2+4sin
2π=2a
2=4,一般地,當(dāng)n=2k-1(k∈N
*)時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182351367169176/SYS201310241823513671691026_DA/4.png)
,
即a
2k+1-a
2k-1=4.所以數(shù)列{a
2k-1}是首項(xiàng)為0、公差為4的等差數(shù)列,
因此a
2k-1=4(k-1).
當(dāng)n=2k(k∈N
*)時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182351367169176/SYS201310241823513671691026_DA/5.png)
,
所以數(shù)列{a
2k}是首項(xiàng)為2、公比為2的等比數(shù)列,因此a
2k=2
k.
故數(shù)列{a
n}的通項(xiàng)公式為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182351367169176/SYS201310241823513671691026_DA/6.png)
(II)由(I)知,S
k=a
1+a
3++a
2k-1=0+4++4(k-1)=2k(k-1),T
k=a
2+a
4++a
2k=2+2
2+2
k=2
k+1-2,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182351367169176/SYS201310241823513671691026_DA/7.png)
于是W
1=0,W
2=1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182351367169176/SYS201310241823513671691026_DA/8.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182351367169176/SYS201310241823513671691026_DA/9.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182351367169176/SYS201310241823513671691026_DA/10.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182351367169176/SYS201310241823513671691026_DA/11.png)
.
下面證明:當(dāng)k≥6時(shí),W
k<1.事實(shí)上,當(dāng)k≥6時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182351367169176/SYS201310241823513671691026_DA/12.png)
,
即W
k+1<W
k.
又W
6<1,所以當(dāng)k≥6時(shí),W
k<1.
故滿足W
k>1的所有k的值為3,4,5.
點(diǎn)評(píng):本題考查數(shù)列的性質(zhì)和應(yīng)用,解題時(shí)要認(rèn)真審題,仔細(xì)解答.