专题三 数列
1.等差数列{a n },{b n }的前n 项和分别为S n ,T n ,且S n T n =7n +45n -3,则使得a n
b n
为整数的正整
数n 的个数是( )
A .3个
B .4个
C .5个
D .6个
2.已知数列{a n }满足:a 1=2,a n +1S n +(S n -1)2=0(n ∈N *),其中S n 为{a n }的前n 项和.若对任意的n 均有(S 1+1)(S 2+1)…(S n +1)≥kn 恒成立,则k 的最大整数值为( )
A .2
B .3
C .4
D .5
3.设S n 是数列{a n }的前n 项和,且a 1=1,(n +1)a n +1=(n -1)S n ,则S n =__________.
4.设数列{a n }的前n 项和为S n ,已知a 1=1,且S n =4-???
?1+2
n a n (n ∈N *),则数列{a n }的通项公式是a n =________.
5.等差数列{a n }的前n 项和为S n ,a 2+a 4=48,a 5=28,S n +30>nλ对一切n ∈N *恒成立,则λ的取值范围为____________.
6.已知等差数列{a n }的公差d ≠0,a 1=0,其前n 项和为S n ,且a 2+2,S 3,S 4成等比数列.
(1)求数列{a n }的通项公式;
(2)若b n =(2n +2)22n +S n +1
,数列{b n }的前n 项和为T n ,求证:T n -2n <3
2.
7.(2018年湖北宜昌部分示范高中教学协作体期中联考)已知正项数列{a n }的前n 项和为S n ,且S n 是1与a n 的等差中项.
(1)求数列{a n }的通项公式;
(2)设T n 为数列????
??2a n a n +1的前n 项和,证明:2
3≤T n <1(n ∈N )*.
8.(2017年天津)已知{a n}为等差数列,前n项和为S n(n∈N*),{b n}是首项为2的等比数列,且公比大于0,b2+b3=12,b3=a4-2a1,S11=11b4.
(1)求{a n}和{b n}的通项公式;
(2)求数列{a2n b n}的前n项和(n∈N*).
9.(2019年浙江)设等差数列{a n}的前n项和为S n,a3=4,a4=S3,数列{b n}满足:对每一个n∈N*,S n+b n,S n+1+b n,S n+2+b n成等比数列.
(1)求数列{a n},{b n}的通项公式;
(2)记C n=a n
2b n,n∈N
*, 证明:C1+C2+…+C n<2 n,n∈N*.
10.(2019年天津)设{a n }是等差数列,{b n }是等比数列.已知a 1=4,b 1=6,b 2=2a 2-2,b 3=2a 3+4.
(1)求{a n }和{b n }的通项公式;
(2)设数列{c n }满足c 1=1,c n =?
????
1,2k b k ,n =2k ,其中k ∈N *. ⅰ)求数列{a 2n (c 2n -1)}的通项公式; ⅱ)求21 n i i i a c =∑(n ∈N * ). 专题三 数列 1.B 解析:a n b n =2a n 2b n =a 1+a 2n -1b 1+b 2n -1=S 2n -1T 2n -1=14n +382n -4=7n +19n -2=7+33 n -2 ,∴n -2=-1或1 或3或11或33.∴n =1或3或5或13或35.当n =3时,S n T n =7n +45 n -3中分母为零,故舍去. 2.B 解析:当n ≥1时,由条件可得S n +1-S n =-()S n -12S n ,从而S n +1-1=S n -1S n ,故 1 S n +1-1-1S n -1=S n S n -1-1S n -1=1,又1S 1-1=1 2-1=1,∴???? ??1S n -1是首项、公差均为1的等差数列,1 S n -1 =n ,S n =n +1n ,依题只须k ≤????(S 1+1)(S 2+1)…(S n +1)n min ,令f (n )=(S 1+1)(S 2+1)…(S n +1)n ,则f (n +1)f (n )=n (S n +1+1)n +1=n (2n +3)(n +1)2 >1,故f (n )min =f (1)=S 1+1 1=3,∴k max =3,选B. 3.2n -1n 解析:∵(n +1)a n +1=(n -1)S n ,∴na n +1+S n +1=nS n ,∴n (S n +1-S n )+S n +1=nS n ,∴(n +1)S n +1nS n =2,∴{nS n }是首项为1,公比为2的等比数列,则nS n =2n -1 ,∴S n =2n - 1 n . 4.n 2 n -1 解析:当n ≥2时,a n =S n -S n -1=????1+2n -1a n -1-????1+2n a n , 则????2+2n a n =????1+2n -1a n -1 ,即a n n =a n -12(n -1), ∴数列???? ??a n n 是首项为1,公比为1 2的等比数列, 则a n n =????12n -1,即a n =n 2n -1 . 5.λ<30 解析:a 5=28,a 2+a 4=a 5+a 1=48,∴a 1=20,d =a 5-a 14=28-20 4 =2,a n = a 1+(n -1)d =2n +18,S n =n (20+2n +18)2=n (n +19),由n (n +19)+30>nλ,得λ< n 2+19n +30 n =n +30n +19,由函数f (x )=x +30 x +19的单调性及f (5)=f (6)=30知,当n =5或n =6时,n + 30 n +19最小,为30,故λ<30. 6.(1)解:由a 1=0得a n =(n -1)d ,S n =n (n -1)d 2 , ∵a 2+2,S 3,S 4成等比数列,∴S 23=(a 2+2)S 4 , 即(3d )2=(d +2)·6d , 整理得3d 2-12d =0,即d 2-4d =0, ∵d ≠0,∴d =4, ∴a n =(n -1)d =4(n -1)=4n -4. (2)证明:由(1)可得S n +1=2n (n +1), ∴b n =(2n +2)22n +2n (n +1)=4(n +1)22n (n +2)=2+2 n (n +2) =2+????1n -1n +2, ∴T n =2n +????1-13+????12-14+…+????1n -1n +2=2n +1+12-1n +1-1 n +2, ∴T n -2n <3 2 . 7.(1)解:n =1时,a 1=1; n ≥2时,4S n -1=(a n -1+1)2,又4S n =(a n +1)2, 两式相减得(a n +a n -1)(a n -a n -1-2)=0. ∵a n >0,∴a n -a n -1=2, ∴{a n }是以1为首项,2为公差的等差数列,即a n =2n -1. (2)证明:2a n a n +1=2(2n -1)(2n +1)=12n -1-1 2n +1 , ∴T n =????1-13+????13-15+…+????12n -1-12n +1=1-12n +1 , ∴T n <1,又∵1a n a n +1 >0,∴T n ≥T 1=2 3, 综上2 3 ≤T n <1成立. 8.解:(1)设等差数列{a n }的公差为d ,等比数列{b n }的公比为q . 由已知b 2+b 3=12,得b 1(q +q 2)=12, 而b 1=2,∴q 2+q -6=0. 又∵q >0,解得q =2. ∴b n =2n . 由b 3=a 4-2a 1,可得3d -a 1=8.① 由S 11=11b 4,可得a 1+5d =16.② 联立①②,解得a 1=1,d =3.由此可得a n =3n -2. ∴{a n }的通项公式为a n =3n -2,{b n }的通项公式为b n =2n . (2)设数列{a 2n b n }的前n 项和为T n ,由a 2n =6n -2, 有T n =4×2+10×22+16×23+…+(6n -2)×2n , 2T n =4×22+10×23+16×24+…+(6n -8)×2n +(6n -2)×2n + 1, 上述两式相减,得-T n =4×2+6×22+6×23+…+6×2n -(6n -2)×2n + 1 =12×(1-2n )1-2 -4-(6n -2)×2n +1=-(3n -4)2n +2-16. ∴T n =(3n -4)2n + 2+16. ∴数列{a 2n b n }的前n 项和为(3n -4)2n + 2+16. 9.(1)解:由题意得????? a 1+2d =4,a 1+3d =3a 1+3×2 2d , 解得????? a 1=0, d =2, 则数列{a n }的通项公式为a n =2n -2. 其前n 项和S n =(0+2n -2)×n 2 =n (n -1). 则n (n -1)+b n ,n (n +1)+b n ,(n +1)(n +2)+b n 成等比数列,即: [n (n +1)+b n ]2=[n (n -1)+b n ]×[(n +1)(n +2)+b n ], ∴n 2(n +1)2+2n (n +1)b n +b 2n =n (n -1)(n +1)(n +2)+(n +1)(n +2)b n +n (n -1)b n +b 2 n , 故b n =n 2(n +1)2-n (n -1)(n +1)(n +2) (n +1)(n +2)+n (n -1)-2n (n +1) =n (n +1). (2)证明:结合(1)中的通项公式可得: C n =a n 2b n =n -1n (n +1)<1n =2n +n <2 n +n -1=2(n -n -1), 则C 1+C 2+…+C n <2(1-0)+2(2-1)+…+2(n -n -1)=2 n . 10.解:(1)设等差数列{a n }的公差为d ,等比数列{}b n 的公比为q . 依题意得????? 6q =6+2d ,6q 2=12+4d ,解得? ???? d =3, q =2, 故a n =4+(n -1)×3=3n +1,b n =6×2n - 1=3×2n . ∴{a n }的通项公式为a n =3n +1,{b n }的通项公式为b n =3×2n . (2)ⅰ)a n 2 (c 2n -1)=a n 2 (b n -1)=(3×2n +1)(3×2n -1)=9×4n -1. ∴数列{a n 2 (c n 2 -1)}的通项公式为a n 2 (c n 2 -1)=9×4n -1. ⅱ)2 1n i a =∑i c i =2 1 [n i =∑a i +a i (c i -1)] = 21n i a =∑i +21 i n i a =∑(c 2i -1) =???? 2n ×4+2n (2n -1)2×3+1n i =∑(9×4i -1) =(3×22n -1+5×2n - 1)+9× 4(1-4n ) 1-4 -n =27×22n -1+5×2n - 1-n -12(n ∈N *).