(新课标)广西201X高考数学二轮复习 专题对点练14 数列与数列不等式的证明及数列中的存在性问题

专题对点练14 数列与数列不等式的证明及数列中的存在性问题1.已知等比数列{a n },a 1=13,公比q=13. (1)S n 为{a n }的前n 项和,证明:S n =1-a a2;(2)设b n =log 3a 1+log 3a 2+…+log 3a n ,求数列{b n }的通项公式.2.已知数列{a n }满足a 1=3,a n+1=3a a -1a a+1. (1)证明:数列{1a a -1}是等差数列,并求{a n }的通项公式;(2)令b n =a 1a 2…a n ,求数列{1a a}的前n 项和S n .3.已知数列{a n }的前n 项和S n =1+λa n ,其中λ≠0. (1)证明{a n }是等比数列,并求其通项公式;(2)若S 5=3132,求λ的值.4.在数列{a n }中,设f (n )=a n ,且f (n )满足f (n+1)-2f (n )=2n (n ∈N *),且a 1=1.(1)设b n =aa 2a -1,证明数列{b n }为等差数列; (2)求数列{a n }的前n 项和S n .5.设数列{a n }的前n 项和为S n ,且(3-m )S n +2ma n =m+3(n ∈N *),其中m 为常数,且m ≠-3. (1)求证:{a n }是等比数列;(2)若数列{a n }的公比q=f (m ),数列{b n }满足b 1=a 1,b n =32f (b n-1)(n ∈N *,n ≥2),求证:{1a a}为等差数列,并求b n .6.已知数列{a n }的前n 项和为S n ,a 1=-2,且满足S n =12a n+1+n+1(n ∈N *). (1)求数列{a n }的通项公式; (2)若b n =log 3(-a n +1),求数列{1aa a a +2}的前n 项和T n ,并求证T n <34.7.(2018天津模拟)已知正项数列{a n },a 1=1,a 2=2,前n 项和为S n ,且满足a a +1a a -1+a a -1a a +1=4a a2a a +1a a -1-2(n ≥2,n ∈N *).(1)求数列{a n }的通项公式; (2)记c n =1aa ·a a +1,数列{c n }的前n 项和为T n ,求证:13≤T n <12.8.已知数列{a n }的前n 项和为S n ,a 1=2,2S n =(n+1)2a n -n 2a n+1,数列{b n }满足b 1=1,b n b n+1=λ·2a a . (1)求数列{a n }的通项公式;(2)是否存在正实数λ,使得{b n }为等比数列?并说明理由.专题对点练14答案1.(1)证明 因为a n =13×(13)a -1=13a,S n =13(1-13a )1-13=1-13a 2,所以S n =1-a a 2.(2)解 b n =log 3a 1+log 3a 2+…+log 3a n =-(1+2+…+n )=-a (a +1)2.所以{b n }的通项公式为b n =-a (a +1)2.2.解 (1)∵a n+1=3a a -1a a+1,∴a n+1-1=3a a -1aa+1-1=2(a a -1)a a +1,∴1a a +1-1=a a+12(a a-1)=1a a-1+12, ∴1aa +1-1−1aa -1=12.∵a 1=3,∴1a 1-1=12,∴数列{1aa -1}是以12为首项,12为公差的等差数列,∴1a a -1=12+12(n-1)=12n ,∴a n =a +2a. (2)∵b n =a 1a 2…a n ,∴b n =31×42×53×…×aa -2×a +1a -1×a +2a=(a +1)(a +2)2,∴1a a=2(a +1)(a +2)=2(1a +1-1a +2),∴S n =2(12−13+13−14+…+1a +1−1a +2)=2(12-1a +2)=aa +2.3.解 (1)由题意得a 1=S 1=1+λa 1,故λ≠1,a 1=11-a,a 1≠0.由S n =1+λa n ,S n+1=1+λa n+1得a n+1=λa n+1-λa n , 即a n+1(λ-1)=λa n .由a 1≠0,λ≠0得a n ≠0,所以a a +1a a=a a -1.因此{a n }是首项为11-a ,公比为aa -1的等比数列, 于是a n =11-a (aa -1)a -1.(2)由(1)得S n =1-(aa -1)a . 由S 5=3132得1-(aa -1)5=3132, 即(a a -1)5=132.解得λ=-1.4.(1)证明 由已知得a n+1=2a n +2n,∴b n+1=a a +12a =2a a +2a2a =aa 2a -1+1=b n +1,∴b n+1-b n =1.又a 1=1,∴b 1=1,∴{b n }是首项为1,公差为1的等差数列.(2)解 由(1)知,b n =aa 2a =n ,∴a n =n·2n-1.∴S n =1+2×21+3×22+…+n·2n-1,2S n =1×21+2×22+…+(n-1)·2n-1+n·2n,两式相减得-S n =1+21+22+…+2n-1-n·2n =2n -1-n·2n =(1-n )2n-1, ∴S n =(n-1)·2n +1.5.证明 (1)由(3-m )S n +2ma n =m+3,得(3-m )S n+1+2ma n+1=m+3, 两式相减,得(3+m )a n+1=2ma n .∵m ≠-3,∴a a +1a a=2aa +3,∴{a n }是等比数列.(2)由(3-m )S n +2ma n =m+3, 得(3-m )S 1+2ma 1=m+3, 即a 1=1,∴b 1=1.∵数列{a n }的公比q=f (m )=2aa +3, ∴当n ≥2时,b n =32f (b n-1)=32·2a a -1a a -1+3,∴b n b n-1+3b n =3b n-1,∴1a a−1aa -1=13.∴{1a a}是以1为首项,13为公差的等差数列,∴1a a=1+a -13=a +23.又1a1=1也符合,∴b n =3a +2. 6.(1)解 ∵S n =12a n+1+n+1(n ∈N *),∴当n=1时,-2=12a 2+2,解得a 2=-8. 当n ≥2时,a n =S n -S n-1=12a n+1+n+1-(12a a +a ), 即a n+1=3a n -2,可得a n+1-1=3(a n -1). 当n=1时,a 2-1=3(a 1-1)=-9,∴数列{a n -1}是等比数列,首项为-3,公比为3. ∴a n -1=-3n ,即a n =1-3n . (2)证明 b n =log 3(-a n +1)=n ,∴1aa a a +2=12(1a -1a +2).∴T n =12[(1-13)+(12-14)+(13-15)+…+(1a -1-1a +1)+(1a -1a +2)]=12(1+12−1a +1−1a +2)<34.∴T n <34.7.(1)解 由aa +1a a -1+a a -1a a +1=4a a2a a +1a a -1-2(n ≥2,n ∈N *),得a a +12+2S n+1S n-1+a a -12=4a a 2,即(S n+1+S n-1)2=(2S n )2.由数列{a n }的各项均为正数,得S n+1+S n-1=2S n ,所以数列{S n }为等差数列.由a 1=1,a 2=2,得S 1=a 1=1,S 2=a 1+a 2=3,则数列{S n }的公差为d=S 2-S 1=2, 所以S n =1+(n-1)×2=2n-1.当n ≥2时,a n =S n -S n-1=(2n-1)-(2n-3)=2,而a 1=1不适合上式,所以数列{a n }的通项公式为a n ={1,a =1,2,a ≥2.(2)证明 由(1)得c n =1a a·a a +1=1(2a -1)(2a +1)=12(12a -1-12a +1),则T n =c 1+c 2+c 3+…+c n =12[(1-13)+(13-15)+(15−17)+…+(12a -1-12a +1)]=12(1-12a +1)<12. 又T n =12(1-12a +1)是关于n 的增函数,则T n ≥T 1=13,因此,13≤T n <12. 8.解 (1)由2S n =(n+1)2a n -n 2a n+1,得2S n-1=n 2a n-1-(n-1)2a n , ∴2a n =(n+1)2a n -n 2a n+1-n 2a n-1+(n-1)2a n ,∴2a n =a n+1+a n-1, ∴数列{a n }为等差数列.∵2S 1=(1+1)2a 1-a 2,∴4=8-a 2. ∴a 2=4.∴d=a 2-a 1=4-2=2. ∴a n =2+2(n-1)=2n.(2)∵b n b n+1=λ·2a a =λ·4n,b 1=1, ∴b 2b 1=4λ,∴b 2=4λ,∴b n+1b n+2=λ·4n+1,∴a a +1a a +2a a a a +1=4, ∴b n+2=4b n ,∴b 3=4b 1=4.若{b n }为等比数列,则a 22=b 3·b 1,∴16λ2=4×1,∴λ=12.故存在正实数λ=12,使得{b n }为等比数列.如有侵权请联系告知删除，感谢你们的配合！。