山东新高考数学二轮复习专题练14求数列的通项及前n项和

专题突破练14求数列的通项及前n项和1.(2019江西宜春高三上学期期末)已知等差数列{a n}的前n项和为S n,且a2+a6=10,S5=20.
(1)求a n与S n;
,求{c n}的前n项和T n.
(2)设数列{c n}满足c n=1
S n-n
2.(2019吉林高中高三上学期期末考试)在递增的等比数列{a n}中,a2=6,且4(a3-a2)=a4-6.
(1)求{a n}的通项公式;
(2)若b n=a n+2n-1,求数列{b n}的前n项和S n.
3.已知数列{a n }满足a 1=12,a n+1=a n 2a n +1
. (1)证明数列{1a n
}是等差数列,并求{a n }的通项公式; (2)若数列{b n }满足b n =12n
·a n ,求数列{b n }的前n 项和S n .
4.(2019辽宁朝阳重点高中高三第四次模拟)已知等差数列{a n }的前n 项和为S n ,满足S 3=12,且a 1,a 2,a 4成等比数列.
(1)求a n 及S n ;
a n n ,数列{
b n}的前n项和为T n,求T n.
(2)设b n=S n·2
5.已知数列{a n}满足a1=1,a2=3,a n+2=3a n+1-2a n(n∈N*).
(1)证明:数列{a n+1-a n}是等比数列;
(2)求数列{a n}的通项公式和前n项和S n.
6.已知等差数列{a n}满足:a n+1>a n,a1=1,该数列的前三项分别加上1,1,3后成等比数列,a n+2log2b n=-1.
(1)求数列{a n },{b n }的通项公式;
(2)求数列{a n ·b n }的前n 项和T n .
7.设S n 是数列{a n }的前n 项和,a n >0,且4S n =a n (a n +2).
(1)求数列{a n }的通项公式;
(2)设b n =1(a n -1)(a n
+1),T n =b 1+b 2+…+b n ,求证:T n <12.
8.(2019山东淄博部分学校高三阶段性诊断考试)已知等比数列{a n}的前n项和为S n(n∈N*),-
2S2,S3,4S4成等差数列,且a2+2a3+a4=1
16
.
(1)求数列{a n}的通项公式;
(2)若b n=-(n+2)log2|a n|,求数列{1
b n
}的前n项和T n.
参考答案
专题突破练14求数列的
通项及前n项和
1.解(1)设等差数列公差为d,
S5=5(a1+a5)
2
=5a3=20,故a3=4,
a2+a6=2a4=10,故a4=5,
∴d=1,a n=a3+d(n-3)=n+1,
易得a1=2,
∴S n=n
2(a1+a n)=n
2
(2+n+1)=n(n+3)
2
.
(2)由(1)知S n =
n (n+3), 则c n =1S n -n =2n 2+n =21n −1n+1, 则T n =21-12+12−13+13−14+…+1n −1n+1=21-1n+1=2n
n+1.
2.解 (1)设公比为q ,由4(a 3-a 2)=a 4-6,得4(6q-6)=6q 2-6,
化简得q 2-4q+3=0,解得q=3或q=1,
因为等比数列{a n }是递增的,所以q=3,a 1=2,所以a n =2×3n-1.
(2)由(1)得b n =2×3n-1+2n-1,
所以S n =(2+6+18+…+2×3n-1)+(1+3+5+…+2n-1),
则S n =2×(1-3n )1-3+n (1+2n -1)2, 所以S n =3n -1+n 2.
3.(1)证明 ∵a n+1=a n 2a n +1
, ∴1a n+1−1a n =2,∴{1
a n }是等差数列, ∴1a n =1a 1
+(n-1)×2=2+2n-2=2n ,即a n =12n . (2)解 ∵b n =1
2n ·a n =2n 2n
, ∴S n =b 1+b 2+…+b n =1+22+
3
22+…+n 2n -1, 则1S n =1+2
22+3
23+…+n
n , 两式相减得12S n =1+1
2+122+1
23+…+12n -1−n n =2(1-1n )−n n ,∴S n =4-2+n 2
n -1.
4.解 (1)设等差数列{a n }的公差为d ,
因为S 3=12,且a 1,a 2,a 4成等比数列,所以有{
S 3=3a 2=12,a 22=a 1a 4, 即{a 1+d =4,(a 1+d )2=a 1(a 1+3d ),
解得{a 1=2,d =2.
所以a n =a 1+(n-1)d=2n ,S n =
n (a 1+a n )2=n 2+n. (2)由(1)可得
b n =S n ·2a n n =n (n+1)·22n n
=(n+1)·4n , 因为数列{b n }的前n 项和为T n ,
所以T n =b 1+b 2+b 3+…+b n =2×4+3×42+4×43+…+(n+1)·4n ,
因此,4T n =2×42+3×43+4×44+…+(n+1)·4n+1,
两式作差,得-3T n =2×4+42+43+44+…+4n -(n+1)·4n+1,
整理得T n =(3n+2)·4n+1-89
. 5.(1)证明 ∵a n+2=3a n+1-2a n (n ∈N *),
∴a n+2-a n+1=2(a n+1-a n )(n ∈N *),
∴a n+2-a
n+1a n+1-a n =2. ∵a 1=1,a 2=3,
∴数列{a n+1-a n }是以a 2-a 1=2为首项,公比为2的等比数列.
(2)解 由(1)得,a n+1-a n =2n (n ∈N *),
∴a n =(a n -a n-1)+(a n-1-a n-2)+…+(a 2-a 1)+a 1=2n-1+2n-2+…+2+1=2n -1,(n ∈N *).
S n =(2-1)+(22-1)+(23-1)+…+(2n -1)=(2+22+23+…+2n )-n=2(1-2n )1-2-n=2n+1-2-n.
6.解 (1)设等差数列{a n }的公差为d ,且d>0,由a 1=1,a 2=1+d ,a 3=1+2d ,分别加上1,1,3后成等比数列,得(2+d )2=2(4+2d ),解得d=2,∴a n =1+(n-1)×2=2n-1.
∵a n +2log 2b n =-1,
∴log 2b n =-n ,即b n =1
2n .
(2)由(1)得a n ·b n =2n -1n .T n =1
21+3
22+5
23+…+2n -1
n ,①
12T n =122+323+524+…+2n -1
2n+1,
② ①-②,得12T n =12+2(122+1
23+124+…+12n )−2n -1
2n+1.
∴T n =1+1-
12n -1
1-12−2n
-12n =3-1
2n -2−2n
-12n =3-2n+3
2n .
7.(1)解 4S n =a n (a n +2),①
当n=1时,4a 1=a 12+2a 1,即a 1=2.
当n ≥2时,4S n-1=a n-1(a n-1+2).
②
由①-②得4a n =a n 2−a n -12+2a n -2a n-1,即2(a n +a n-1)=(a n +a n-1)·(a n -a n-1). ∵a n >0,∴a n -a n-1=2,
∴a n =2+2(n-1)=2n.
(2)证明 ∵b n =1
(a n -1)(a n +1)
=1
(2n -1)(2n+1)
=12(12n -1-12n+1),
∴T n=b1+b2+…+b n=1
21-1
3
+1
3
−1
5
+…+1
2n-1
−1
2n+1
)=1
2
1-1
2n+1
<1
2
.
8.解(1)设等比数列{a n}的公比为q.
由-2S2,S3,4S4成等差数列知,
2S3=-2S2+4S4,
所以2a4=-a3,即q=-1
2
.
又a2+2a3+a4=1,
所以a1q+2a1q2+a1q3=1
16
,
所以a1=-1
2
.
所以等差数列{a n}的通项公式a n=-1
2
n.
(2)由(1)知
b n=-(n+2)log2-1
2
n=n(n+2),
所以1
b n =1
n(n+2)
=1
2
1
n
−1
n+2
.
所以数列1
b n
的前n项和:
T n=1
21-1
3
+1
2
−1
4
+1
3
−1
5
+…+1
n-1
−1
n+1
+1
n
−1
n+2
=1 21+1
2
−1
n+1
−1
n+2
=3
4
−2n+3
2(n+1)(n+2)
.所以数列1
b n
的前n项和T n=3
4
−
2n+3
2(n+1)(n+2)
.。