2019年高考数学(文科)二轮复习对点练：四数列专题对点练14(含答案)

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=,公比q=.
(1)S n 为{a n }的前n 项和,证明:S n =;
1-a n
2(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 n -1
a n +1(1)证明:数列是等差数列,并求{a n
}的通项公式;
{
1a n -1}
(2)令b n =a 1a 2…a n
,求数列的前n 项和S n
.
{1b 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=,求λ的值.
31
324.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 =,证明数列{b n }为等差数列;a n
2n -1
(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 =f (b n-1)(n ∈N *,n ≥2),求证:为等差数列,并求b n .
{1b n }
6.已知数列{a n }的前n 项和为S n ,a 1=-2,且满足S n =a 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),求数列的前n 项和T n ,并求证T n <.
{1b n b n +2}
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-
S n +1
S n -1
+
S n
-1
S n +1
=
4S 2
n
S n +1S n
-12(n ≥2,n ∈N *).
(1)求数列{a n }的通项公式;
(2)记c n
=,数列{c n
}的前n 项和为T n
,求证:≤T n
<.
1S n ·S n +18.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
n (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 =
,S n =,
13×(1
3)
n -
1=13n 13(1-13n )
1-13=
1-13n
2所以S n =
.
1-a n
2
(2)解 b n =log 3a 1+log 3a 2+…+log 3a n =-(1+2+…+n )=-.
n (n +1)
2所以{b n }的通项公式为b n =-.
n (n +1)
22.解 (1)∵a n+1=,∴a n+1-1=-1=,3a n -1a n +13a n -1a n +12(a n -1)
a n +1∴,
1a n +1-1=a n +12(a n -1)=1a n -1+
12∴.
1
a n +1-1‒
1
a n -1
=
12∵a 1
=3,∴,
1a 1-1=
1
2∴数列是以为首项,为公差的等差数列,∴(n-1)= n ,∴a n
=.
{
1
a n -1}
1
a n -1=
12+
12n +2n (2)∵b n =a 1a 2…a n ,
∴b n =×…×,31×42×53n n -2×n +1n -1×n +2n =(n +1)(n +2)2∴=2,
1b n =2(n +1)(n +2)(
1n +1-1n +2)
∴S n =2+…+=2.
(12‒13+13‒141n +1‒1n +2)(12-1n +2)
=n n +23.解 (1)由题意得a 1=S 1=1+λa 1,故λ≠1,a 1=,a 1≠0.
1
1-λ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 n +1a n
=
λ
λ-1因此{a n }是首项为,公比为的等比数列,
11-λλλ-1于是a n =.
11-λ(λλ-1
)
n -1(2)由(1)得S n =1-.
(λ
λ-1)
n
由S 5=得1-,3132(λλ-1)
5=
31
32即
.(λλ-1)
5=
1
32解得λ=-1.
4.(1)证明 由已知得a n+1=2a n +2n ,
∴b n+1=+1=b n +1,a n +12
n
=2a n +2n 2n =a n
2n -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 ==n ,a n
2n -1
∴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 n +1a n
=
2m
m +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 )=,
2m
m +3∴当n ≥2时,b n =f (b n-1)=,32·2b n -1b n -1+3∴b n b n-1+3b n =3b n-1
,∴.
1b n ‒1b n -1=
13∴是以1为首项,为公差的等差数列,∴=1+.{1b n }
1b n n -13=
n +2
3又=1也符合,∴b n
=.
1b 13n +26.(1)解 ∵S n =a n+1+n+1(n ∈N *),∴当n=1时,-2=a 2+2,解得a 2=-8.
当n ≥2时,a n =S n -S n-1=a n+1+n+1-,
(1
2a n +n )
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 ,
∴.1b n b n +2=12(1n -1
n +2)∴T n =+…+.∴12
[(
1-13)+(12-14)+(
13-15)(
1n -1-1n +1)+(1n -1n +2)]=12(1+12‒1n +1‒1n +2)
<34T n <.
7.(1)解 由-2(n ≥2,n ∈N *),得+2S n+1S n-1+=4,S n +1S n -1
+S n -1S n +1=4S 2
n
S n +1S n -1S 2n +1S 2n -1S 2
n 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,n =1,
2,n ≥2.
(2)证明 由(1)得c n =,1S n ·S n +1=1(2n -1)(2n +1)=12(12n -1-1
2n +1)
则T n =c 1+c 2+c 3+…+c n =+…+1-.12
[(1-13)+(
13-15)+(15‒17)
(
12n -1-12n +1)]=12( 12n +1)
<12又T n =是关于n 的增函数,则T n ≥T 1=,因此,≤T n <.12
(1-12n +1)
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·=λ·4n ,b 1=1,2a
n
∴b 2b 1=4λ,∴b 2=4λ,
∴b n+1b n+2=λ·4n+1,∴=4,b n +1b n +2
b n b n +1∴b n+2=4b n ,∴b 3=4b 1=4.
若{b n }为等比数列,则=b 3·b 1,∴16λ2=4×1,∴λ=.b 2
2故存在正实数λ=,使得{b n }为等比数列.。