【答案】
分析:根據(jù)數(shù)列為無(wú)窮等比數(shù)列,且所有項(xiàng)的和為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/0.png)
,得到極限存在,即公比q大于等于-1小于等于1,且不為0,當(dāng)官公比q等于1時(shí),數(shù)列為常數(shù)列,利用首項(xiàng)a
1表示出數(shù)列的前n項(xiàng)的和,解出a
1,當(dāng)n趨于無(wú)窮大時(shí)a
1趨于0,得到a
1大于0,當(dāng)q大于等于-1小于1時(shí),利用等比數(shù)列的前n項(xiàng)和公式表示出s
n,當(dāng)n趨于無(wú)窮大時(shí),q
n趨于0,得到s
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/2.png)
,解出a
1,根據(jù)當(dāng)q=-1時(shí),a
1取得最大值,即可解出a
1的取值范圍,同時(shí)因?yàn)楣萹不為0,得到a
1不等于
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/3.png)
,綜上,寫(xiě)出a
1的取值范圍即可.
解答:解:因?yàn)閿?shù)列{a
n}為無(wú)窮等比數(shù)列,且其所有項(xiàng)的和為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/4.png)
,即其極限存在,
故可知|q|≤1且q≠0,即-1≤q≤1且q≠0,
當(dāng)q=1時(shí),無(wú)窮等比數(shù)列{a
n}為常數(shù)列,設(shè)s
n為其所有項(xiàng)之和,則s
n=na
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/5.png)
,
即a
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/6.png)
,當(dāng)n→+∞時(shí),a
1→0,即a
1>0;
當(dāng)-1≤q<1時(shí),s
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/7.png)
,當(dāng)n→+∞時(shí),q
n→0,于是有s
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/9.png)
,
即a
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/10.png)
(1-q),當(dāng)q=-1時(shí),a
1最大,所以得到0<a
1≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/11.png)
,
又q≠0,得到a
1≠
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/12.png)
,
綜上,a
1的范圍是(0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/13.png)
)∪(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/14.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/15.png)
).
故答案為:(0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/16.png)
)∪(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/17.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/18.png)
)
點(diǎn)評(píng):此題考查學(xué)生靈活運(yùn)用等比數(shù)列的前n項(xiàng)和公式化簡(jiǎn)求值,要求學(xué)生會(huì)利用極限思想解決實(shí)際問(wèn)題,是一道中檔題.學(xué)生求a
1范圍的時(shí)候注意q不為0這個(gè)條件得到a
1≠
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379968065/SYS201310241812103799680008_DA/19.png)
.