全国统考2022高考数学一轮复习高考大题专项三数列理含解析北师大版

高考大题专项(三) 数列
1.(2020广东天河区模拟)已知S n为数列{a n}的前n项和,且a1<2,a n>0,6S n=a n2+3a n+2,n∈N+.
(1)求数列{a n}的通项公式;
(2)若任意n∈N+,b n=(-1)n a n2,求数列{b n}的前2n项和T2n.
2.(2020湖南郴州二模,文17)设等差数列{a n}的公差为d(d>1),前n项和为S n,等比数列{b n}的公比为q.已知b1=a1,b2=3,2q=3d,S10=100.
(1)求数列{a n},{b n}的通项公式;
(2)记c n=a n·b n,求数列{c n}的前n项和T n.
3.(2020安徽合肥4月质检二,理17)已知等差数列{a n}的前n项和为S n,a2=1,S7=14,数列{b n}满足
b1·b2·b3·…·b n=2n2+n 2.
(1)求数列{a n}和{b n}的通项公式;
(2)若数列{c n}满足c n=b n cos(a nπ),求数列{c n}的前2n项和T2n.
4.(2020山西长治二模)S n 为等比数列{a n }的前n 项和,已知a 4=9a 2,S 3=13,且公比q>0. (1)求a n 及S n .
(2)是否存在常数λ,使得数列{S n +λ}是等比数列?若存在,求出λ的值;若不存在,请说明理由.
5.(2020天津,19)已知{a n }为等差数列,{b n }为等比数列,a 1=b 1=1,a 5=5(a 4-a 3),b 5=4(b 4-b 3). (1)求{a n }和{b n }的通项公式;
(2)记{a n }的前n 项和为S n ,求证:S n S n+2<S n+12
(n ∈N +);
(3)对任意的正整数n ,设c n ={(3a n -2)b n
a n
a n+2
,n 为奇数,
a n -1
b n+1
,n 为偶数.
求数列{c n }的前2n 项和.
参考答案
高考大题专项(三) 数列
1.解(1)当n=1时,6a 1=a 12
+3a 1+2,且a 1<2,解得a 1=1.
当n ≥2时,6a n =6S n -6S n-1=a n 2
+3a n +2-(a n -12+3a n-1+2).
化简,得(a n +a n-1)(a n -a n-1-3)=0,因为a n >0,所以a n -a n-1=3,
所以数列{a n }是首项为1,公差为3的等差数列,所以a n =1+3(n-1)=3n-2.
(2)b n =(-1)n a n 2=(-1)n (3n-2)2.
所以b 2n-1+b 2n =-(6n-5)2+(6n-2)2=36n-21.
所以数列{b n }的前2n 项的和T 2n =36×(1+2+…+n )-21n=36×n (n+1)2
-21n=18n 2-3n.
2.解(1)由题意,得{S 10=10a 1+45d =100,b 2=b 1q =3,
将b 1=a 1,q=3
2d 代入上式,
可得{2a 1+9d =20,a 1d =2,解得{a 1=9,d =29
(舍去),或{a 1=1,
d =2.
∴数列{a n }的通项公式为a n =1+2(n-1)=2n-1,n ∈N +. ∴b 1=a 1=1,q=3
2
d=3
2×2=3,
∴数列{b n }的通项公式为b n =1×3n-1=3n-1,n ∈N +. (2)由(1)知,c n =a n ·b n =(2n-1)·3n-1,
∴T n =c 1+c 2+c 3+…+c n =1×1+3×3+5×32+…+(2n-1)·3n-1, ① 3T n =1×3+3×32+…+(2n-3)·3n-1+(2n-1)·3n , ②
①-②,得-2T n
=1+2×3+2×32+…+2·3n-1-(2n-1)·3n =1+2×(3+32
+…+3n-1)-(2n-1)·3n =1+2×3-3n 1-3
-(2n-
1)·3n =-(2n-2)·3n -2,
∴T n =(n-1)·3n +1.
3.解(1)设数列{a n }的公差为d ,由a 2=1,S 7=14,得{a 1+d =1,
7a 1+21d =14.解得{a 1=1
2,d =12,
所以a n =n
2
.∵b 1·b 2·b 3·…·b n =2
n 2+n
2
=2
n (n+1)2
,∴b 1·b 2·b 3·…·b n-1=2
n (n -1)
2
(n ≥2),
两式相除,得b n =2n (n ≥2).
当n=1时,b 1=2适合上式.∴b n =2n . (2)∵c n =b n cos(a n π)=2n cos (n
2π),
∴T 2n =2cos π
2+22cos π+23cos 3π
2+24cos(2π)+…+22n-1cos (2n -1)π2
+22n cos(n π)
则T 2n
=22cos π+24cos(2π)+26cos(3π)+…+22n cos(n π)=-22+24-26+…+(-1)n ·22n =-4×[1-(-4)n
]1+4
=-4+(-4)
n+1
5
.
4.解(1)由题意可得{
a 1q 3=9a 1q ,
a 1(1-q 3)1-q
=13,q >0,
解得{a 1=1,
q =3,
所以a n
=3n-1,S
n =
1-3n 1-3
=3n -12
.
(2)假设存在常数λ,使得数列{S n +λ}是等比数列,因为S 1+λ=λ+1,S 2+λ=λ+4,S 3+λ=λ+13,所以(λ+4)2=(λ+1)(λ+13),解得λ=12,此时S n +12=1
2×3n
,则
S n+1+
12
S n +
12
=3,故存在常数λ=12,使得数列{S n +1
2}是等比
数列.
5.(1)解设等差数列{a n }的公差为d ,等比数列{b n }的公比为q.由a 1=1,a 5=5(a 4-a 3),可得d=1,从而{a n }的通项公式为a n =n.由b 1=1,b 5=4(b 4-b 3),又q ≠0,可得q 2-4q+4=0,解得q=2,从而{b n }的通项公式为b n =2n-1.
(2)证明由(1)可得S n =
n (n+1)2
,故S n S n+2=14
n (n+1)(n+2)(n+3),S n+12=1
4
(n+1)2(n+2)2,从而S n S n+2-S n+12
=-12
(n+1)(n+2)<0,所以S n S n+2<S n+12
.
(3)解当n 为奇数时,c n =
(3a n -2)b n a n a n+2=
(3n -2)2n -1n (n+2)
=
2n+1n+2
−
2n -1n
;当n 为偶数时,c n =
a n -1
b n+1
=
n -12n
.对任意的正整数
n ,有∑k=1n
c 2k-1=∑
k=1
n
22k
2k+1
−22k -22k -1=22n
2n+1-1,
∑k=1
n
c 2k =∑
k=1
n 2k -1
4k
=14
+
342+543
+…+
2n -14n
. ① 由①得14∑k=1n
c 2k =
1
4
2+
34
3+…+
2n -34n +
2n -14n+1.
②
由①②得34∑k=1n
c 2k =14+2
42+…+2
4n −2n -1
4n+1=
24(1-1
4
n )1-14
−1
4−2n -1
4n+1,从而得∑k=1
n
c 2k =5
9−6n+5
9×4n .
因此,∑k=1
2n
c k =∑k=1
n
c 2k-1+∑k=1
n
c 2k =4n
2n+1−6n+5
9×4n −4
9.所以,数列{c n }的前2n 项和为4n
2n+1−6n+5
9×4n −4
9.。