【答案】
分析:(1)根據(jù)由a
n+a
n+1=2
n,代入
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181340271182387/SYS201310241813402711823019_DA/0.png)
化簡(jiǎn)整理可知結(jié)果為常數(shù),故可根據(jù)等比數(shù)列的定義判斷數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181340271182387/SYS201310241813402711823019_DA/1.png)
是等比數(shù)列,公比為-1,首項(xiàng)為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181340271182387/SYS201310241813402711823019_DA/2.png)
,進(jìn)而可得等比數(shù)列的通項(xiàng)公式.
(2)先假設(shè)在數(shù)列{b
n}中,存在連續(xù)三項(xiàng)b
k-1,b
k,b
k+1(k∈N
*,k≥2)成等差數(shù)列,再根據(jù)等差中項(xiàng)的性質(zhì)可得b
k-1+b
k+1=2b
k,再通過(1)求得的a
n,求得b
n代入整理得2
k-1=4(-1)
k-1,分k為奇數(shù)和偶數(shù)兩種情況分別討論,進(jìn)而求得k.
(3)①要使b
1,b
r,b
s成等差數(shù)列,只需b
1+b
s=2b
r,代入bn的通項(xiàng)公式整理得2
s-2
r+1=(-1)
s-2(-1)
r-3,分情況討論,若s=r+1,當(dāng)s為不小于4的正偶數(shù),且s=r+1時(shí)符合條件;若s≥r+2時(shí)根據(jù)r的范圍推斷等式不成立.綜合可得答案.
②假設(shè)存在滿足條件1<r<s<t的正整數(shù)r,s,t,使得b
1,b
r,b
s,b
t成等差數(shù)列.首先找到成等差數(shù)列的3項(xiàng),根據(jù)b
t=b
2n+d和b
t=2
t-(-1)
t,進(jìn)而整理得2
t-3×2
2n-1=(-1)
t-3.由于左端大于等于8;右邊小于等于-2,進(jìn)而推斷等式不可能成立,最后綜和即可得出結(jié)論.
解答:解:(1)證明:由a
n+a
n+1=2
n,得a
n+1=2
n-a
n,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181340271182387/SYS201310241813402711823019_DA/3.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181340271182387/SYS201310241813402711823019_DA/4.png)
,
∴數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181340271182387/SYS201310241813402711823019_DA/5.png)
是首項(xiàng)為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181340271182387/SYS201310241813402711823019_DA/6.png)
,公比為-1的等比數(shù)列.
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181340271182387/SYS201310241813402711823019_DA/7.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181340271182387/SYS201310241813402711823019_DA/8.png)
,
∴b
n=2
n-(-1)
n(2)解:假設(shè)在數(shù)列{b
n}中,存在連續(xù)三項(xiàng)b
k-1,b
k,b
k+1(k∈N
*,k≥2)成等差數(shù)列,
則b
k-1+b
k+1=2b
k,即[2
k-1-(-1)
k-1]+[2
k+1-(-1)
k+1]=2[2
k-(-1)
k],
即2
k-1=4(-1)
k-1①若k為偶數(shù),則2
k-1>0,4(-1)
k-1=-4<0,所以,不存在偶數(shù)k,使得b
k-1,b
k,b
k+1成等差數(shù)列.
②若k為奇數(shù),則k≥3,∴2
k-1≥4,而4(-1)
k-1=4,所以,當(dāng)且僅當(dāng)k=3時(shí),b
k-1,b
k,b
k+1成等差數(shù)列.
綜上所述,在數(shù)列{b
n}中,有且僅有連續(xù)三項(xiàng)b
2,b
3,b
4成等差數(shù)列.
(3)①證明:要使b
1,b
r,b
s成等差數(shù)列,只需b
1+b
s=2b
r,即3+2
s-(-1)
s=2[2
r-(-1)
r],
即2
s-2
r+1=(-1)
s-2(-1)
r-3,①
(�。┤魋=r+1,在①式中,左端2
s-2
r+1=0,右端(-1)
s-2(-1)
r-3=(-1)
s+2(-1)
s-3=3(-1)
s-3,
要使①式成立,當(dāng)且僅當(dāng)s為偶數(shù)時(shí)成立.又s>r>1,且s,r為正整數(shù),
所以,當(dāng)s為不小于4的正偶數(shù),且s=r+1時(shí),b
1,b
r,b
s成等差數(shù)列.
(ⅱ)若s≥r+2時(shí),在①式中,左端2
s-2
r+1≥2
r+2-2
r+1=2
r+1,
由(2)可知,r≥3,∴r+1≥4,
∴2
s-2
r+1≥16;右端(-1)
s-2(-1)
r-3≤0(當(dāng)且僅當(dāng)s為偶數(shù)、r為奇數(shù)時(shí)取“=”),
∴當(dāng)s≥r+2時(shí),b
1,b
r,b
s不成等差數(shù)列.
綜上所述,存在不小于4的正偶數(shù)s,且s=r+1,使得b
1,b
r,b
s成等差數(shù)列.
②假設(shè)存在滿足條件1<r<s<t的正整數(shù)r,s,t,使得b
1,b
r,b
s,b
t成等差數(shù)列.
首先找到成等差數(shù)列的3項(xiàng):由第(3)小題第①問,可知,b
1,b
2n-1,b
2n(n∈N*,且n≥2)成等差數(shù)列,
其公差d=b
2n-b
2n-1=[2
2n-(-1)
2n]-[2
2n-1-(-1)
2n-1]=2
2n-1-2,
∴b
t=b
2n+d=2
2n-(-1)
2n+2
2n-1-2=3×2
2n-1-3.
又b
t=2
t-(-1)
t,
∴3×2
2n-1-3=2
t-(-1)
t,
即2
t-3×2
2n-1=(-1)
t-3.②
∵t>2n>2n-1,∴t≥2n+1,
∴②式的左端2
t-3×2
2n-1≥2
2n+1-3×2
2n-1=2
2n-1≥8,
而②式的右端(-1)
t-3≤-2,
∴②式不成立.
綜上所述,不存在滿足條件1<r<s<t的正整數(shù)r,s,t,使得b
1,b
r,b
s,b
t成等差數(shù)列.
點(diǎn)評(píng):本題主要考查等比數(shù)列的判定和等差數(shù)列的應(yīng)用.?dāng)?shù)比數(shù)列常與對(duì)數(shù)函數(shù)、不等式等問題一塊考查,故應(yīng)綜合掌握.